ENERGY. Conceptual Questions Kinetic energy depends on speed. Potential energy depends on position.

Transcription

1 ENERGY 0 Conceptual Questions 0.. Kinetic energy depends on speed. Potential energy depends on position. 0.. No, kinetic energy can never be negative. Kinetic energy is energy of motion. Motion may stop, but it can t be negative. Speed has no direction and cannot be negative. Yes, gravitational potential energy can be negative. Potential energy depends upon position, which can be positive or negative We must calculate the new kinetic energy and compare it to the original value. Originally, velocity of 3v, K = m(3 v) = 9 mv = 9K. The kinetic energy increases by a factor of 9. K = mv. With a 0.4. We have Since m A = mb, K A = 8K B mv A A= 8 mv B B mb va= 8 mv B B va = 4 v B 0.5. Conservation of energy tells us that U i = K f, since the car starts at rest. Originally, this means that in rolling down a track of height h, mgh = mv0 To go twice as fast at the bottom, we must find the height h such that mgh = m ( v0 ) mgh = 4 mv0. So h = 4h. You must increase the track height by a factor of 4. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. 0-

2 0- Chapter va = vb = vc. They each start with the same kinetic energy and they each have the same change in potential energy, so they end with the same kinetic energy and, thus, the same speed va = vb = vc. The balls start off with the same kinetic energy and have the same change in potential energy, so their final kinetic energy is the same (a) We identify the equilibrium position s e = 0 cm. At s = cm, s= s se = cm 0 cm = cm, and F = F = k s = k( cm). To get F sp = 3 F, we must have 3 F = k( s ), which means ( s ) = 3 cm. So the sp spring must have length 0 cm + 3 cm = 3 cm. (b) Note that the direction of the force is reversed when the spring is compressed. To get F sp = F, we must have F = k( s ), which means ( s ) = cm. So the spring must have length 0 cm cm = 8 cm Note that Carlos takes the place of the wall, and that the force on the spring is still 00 N. The spring still stretches 0 cm ( Us) d> ( Us) c> ( Us) b= ( Us) a. Us = k ( s ). Increasing the stretch by a factor of increases the stored energy by a factor of 4. Doubling k doubles the stored energy. 0.. The original spring stores energy ( 0 cm) U = k.. For a spring with spring constant k = k, U = k ( s) = ( k)( s) If U = U, then k(. 0 cm) = ( k)( s) s = cm = 0. 7 cm 0.. Energy conservation tells us that the initial potential energy stored in the spring is equal to the final kinetic energy of the ball. k( s) = mv 0 When the spring is compressed twice as far, k( s) = m( v0) So the ball speed increases by a factor of (a) At x = 6 m the particle has the most kinetic energy. The kinetic energy is the difference between the total energy (TE) and potential energy (PE) curves. At x = 3 m the particle s speed is locally a maximum, but is not as fast as at x = 6 m. (b) The turning points for the particle with total energy (TE) shown are at x = m and x = 8 m. (c) The particle could remain at rest in stable equilibrium at x = 3 m and x = 6 m. The particle could also remain at rest in unstable equilibrium at x = m and x = 4 m The problem can be divided into three parts: () from when the first ball is released and to just before it hits the stationary ball, () the two balls collide, and (3) the two balls swing up together just after the collision to their highest point. Energy is conserved in parts () and (3) as the balls swing like pendulums, but during the collision in Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

3 Energy 0-3 part () momentum is conserved but energy is not. So both energy and momentum conservation are each separately used as you work through each part of the problem. Exercises and Problems Section 0. Kinetic Energy and Gravitational Potential Energy 0.. Model: We will use the particle model for the bullet (B) and the running student (S). Solve: For the bullet, KB= mv (0 00 kg)(500 m/s) 50 J B B=. = For the running student, KS= mv (75 kg)(5 5 m/s) 06 J S S=. = Thus, the bullet has the larger kinetic energy. Assess: Kinetic energy depends not only on mass but also on the square of the velocity. The above calculation shows this dependence. Although the mass of the bullet is 7500 times smaller than the mass of the student, its speed is more than 90 times larger. 0.. Model: Model the hiker as a particle. The origin of the coordinate system chosen for this problem is at sea level so that the hiker s position in Death Valley is y 0 =. 85m. Solve: The hiker s change in potential energy from the bottom of Death Valley to the top of Mt. Whitney is U = U U = mgy mgy = mg( y y ) Assess: Note that gf gi f i f i 6 = (65 kg)(9. 8 m/s )[440 m ( 85 m)] =. 9 0 J U is independent of the origin of the coordinate system. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

4 0-4 Chapter Model: Model the compact car (C) and the truck (T) as particles. Solve: For the kinetic energy of the compact car and the kinetic energy of the truck to be equal, m 0,000 kg K = K mv = mv v = v = (5 km/h) = km/h 000 kg T C T C C T T C T mc Assess: A smaller mass needs a greater velocity for its kinetic energy to be the same as that of a larger mass Model: Model the car (C) as a particle. This is an example of free fall, and therefore the sum of kinetic and potential energy does not change as the car falls. Solve: (a) The kinetic energy of the car is 5 KC= mv (500 kg)(30 m/s) J C C= =. 5 The car s kinetic energy is J. (b) Let us relabel K C as K f and place our coordinate system at y f = 0 m so that the car s potential energy U gf is zero, its velocity is v f, and its kinetic energy is K f. At position y i, vi = 0 m/s or Ki = 0 J, and the only energy the car has is Ugi = mgyi. Since the sum K + Ug is unchanged by motion, Kf + Ugf = Ki + Ugi.This means (c) From part (b), K + mgy = K + mgy K + 0 = K + mgy f f i i f i i 5 ( Kf Ki) ( J 0 J) yi = = = 46 m mg (500 kg)(9.8 m/s ) f mv i f mv i f i ( K K ) ( v v ) yi = = = mg mg g Free fall does not depend upon the mass Model: This is a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball rises and falls. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

5 Energy 0-5 The figure shows a ball s before-and-after pictorial representation for the three situations in parts (a), (b), and (c). Solve: The quantity K + Ug is the same during free fall: Kf + Ugf = Ki + Ugi.We have (a) mv mgy mv mgy + = y = ( v v )/ g = [(0 m/s) (0 m/s) ]/( 9. 8 m/s ) = 5. 0 m 5. m is therefore the maximum height of the ball above the window. This is 5. m above the ground. (b) mv + mgy = mv0 + mgy0 Since y = y0 = 0, we get for the magnitudes v = v0 = 0 m/s. (c) mv3 + mgy3 = mv0 + mgy0 v3 + gy3 = v0 + gy0 v3 = v0 + g( y0 y3) 3 (0 m/s) (9.8 m/s )[0 m ( 0 m)] 49 m /s v = + = This means the magnitude of v 3 is equal to m/s. Assess: Note that the ball s speed as it passes the window on its way down is the same as the speed with which it was tossed up, but in the opposite direction Model: This is a problem of free fall. The sum of the kinetic and gravitational potential energy for the ball, considered as a particle, does not change during its motion. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

6 0-6 Chapter 0 The figure shows the ball s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: The quantity K + Ug is the same during free fall. Thus, Kf + Ugf = Ki + Ugi. (a) (b) mv + mgy= mv0 + mgy0 v0 = v + g( y y0) 0 v0 v = (0 m/s) + (9. 8 m/s )(0 m. 5 m) = m /s =. 9 m/s 3 m/s mv + mgy = mv0 + mgy0 v = v0 + g( y0 y) v v = m /s + (9. 8 m/s )(. 5 m 0 m) = 4 m/s Assess: An increase in speed from.9 m/s to 4.0 m/s as the ball falls through a distance of.5 m is reasonable. Also, note that mass does not appear in the calculations that involve free fall Model: Model the mother and the son as particles. mmother = 4 mson. Solve: The energy conservation equation Kmother = Kson is vson mmothervmother = msonvson (4 mson ) vmother = msonvson =. 0 v Assess: The result vson = vmother, combined with the fact that m son = mmother, is a consequence of the way kinetic energy is defined: It is directly proportional to the mass and to the square of the speed. Section 0.3 A Closer Look at Gravitational Potential Energy 0.8. Model: Model the skateboarder as a particle. Assuming that the track offers no rolling friction, the sum of the skateboarder s kinetic and gravitational potential energy does not change during his rolling motion. 4 mother The vertical displacement of the skateboarder is equal to the radius of the track. Solve: The quantity K + U is the same at the upper edge of the quarter-pipe track as it was at the bottom. The energy conservation equation Kf + Ugf = Ki + Ugi is g mvf + mgyf = mvi + mgyi vi = vf + g( yf yi) i v = (0 m/s) + (9. 8 m/s )(3. 0 m 0 m) = m /s v = 7. 7 m/s Assess: Note that we did not need to know the skateboarder s mass, as is the case with free-fall motion Model: Model the puck as a particle. Since the ramp is frictionless, the sum of the puck s kinetic and gravitational potential energy does not change during its sliding motion. i Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

7 Energy 0-7 Solve: The quantity K + U is the same at the top of the ramp as it was at the bottom. The energy conservation equation Kf + Ugf = Ki + Ugi is g mvf + mgyf = mvi + mgyi vi = vf + g( yf yi) i v = (0 m/s) + (9. 8 m/s )(. 03 m 0 m) = 0. m /s v = 4. 5 m/s Assess: An initial push with a speed of 4.5 m/s 0 mph to cover a distance of 3.0 m up a 0 ramp seems reasonable Model: In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change as the pendulum swings from one side to the other. i Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

8 0-8 Chapter 0 The figure shows the pendulum s before-and-after pictorial representation for the two situations described in parts (a) and (b). Solve: (a) The quantity K + U is the same at the lowest point of the trajectory as it was at the highest point. Thus, g K + U g = K 0 + U g0 means mv + mgy = mv + mgy v + gy = v + gy (0 m) (0 m/s) 0 0 v + g = + gy v = gy From the pictorial representation, we find that y0 = L Lcos30. Thus, v = gl( cos30 ) = (9. 8 m/s )(0. 75 m)( cos30 ) =. 403 m/s The speed at the lowest point is.4 m/s. (b) Since the quantity K + Ug does not change, K + Ug = K+ Ug. We have Since y = L Lcos θ, we obtain mv + mgy = mv + mgy y = ( v v)/g y = [(. 403 m/s) (0 m/s) ]/( 9. 8 m/s ) = m L y (0. 75 m) (0. 0 m) cosθ = = = θ = cos ( ) = 30 L (0. 75 m) That is, the pendulum swings to the other side by 30. Assess: The swing angle is the same on either side of the rest position. This result is a consequence of the fact that the sum of the kinetic and gravitational potential energy does not change. This is shown as well in the energy bar chart in the figure. 0.. Model: Model the child and swing as a particle, and assume the chain to be massless. In the absence of frictional and air-drag effects, the sum of the kinetic and gravitational potential energy does not change during the swing s motion. Solve: The quantity K + U is the same at the highest point of the swing as it is at the lowest point. That is, g K0 + Ug0 = K + Ug.It is clear from this equation that maximum kinetic energy occurs where the gravitational potential energy is the least. This is the case at the lowest position of the swing. At this position, the speed of the swing and child will also be maximum. The above equation is mv0 + mgy0 = mv + mgy v = v0 + g( y0 y) (0 m/s) ( 0 0 m) 0 v = + g y v = gy Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

9 Energy 0-9 We see from the pictorial representation that y = L Lcos45 = (3. 0 m) (3. 0 m)cos45 = m 0 v = gy = (9. 8 m/s )( m) = 4. m/s 0 Assess: We did not need to know the swing s or the child s mass. Also, a maximum speed of 4. m/s is reasonable. 0.. Model: Model the car as a particle with zero rolling friction and no air resistance. The sum of the kinetic and gravitational potential energy, therefore, does not change during the car s motion. Solve: The initial energy of the car is 5 K0 + Ug0 = mv0 + mgy0 = (500 kg)(0. 0 m/s) + (500 kg)(9. 8 m/s )(0 m) =. 0 J The car increases its height to 5 m at the gas station. The conservation of energy equation K0 + Ug0 = K + Ug is J = mv + mgy. 0 J = (500 kg) v + (500 kg)(9. 8 m/s )(5 m) v =. 4 m/s Assess: A lower speed at the gas station is reasonable because the car has decreased its kinetic energy and increased its potential energy compared to its starting values. Section 0.4 Restoring Forces and Hooke s Law 0.3. Model: Assume that the spring is ideal and obeys Hooke s law. According to Hooke s law, the spring force acting on a mass (m) attached to the end of a spring is given as Fsp = k x, where x is the change in length of the spring. If the mass m is at rest, then F sp is also equal to the gravitational force F G Solve: = mg. We have Fsp = k x = mg. We want a 0.00 kg mass to give x = m. This means k = mg/ x = (0. 00 kg)(9. 8 N/m)/(0. 00 m) = 98 N/m 0.4. Model: Assume an ideal spring that obeys Hooke s law. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

10 0-0 Chapter 0 Solve: (a) The spring force on the.0 kg mass is F sp =. k y Notice that y is negative, so F sp is positive. This force is equal to mg, because the.0 kg mass is at rest. We have k y = mg.solving for k: The spring constant is N/m. (b) Again using k y = mg: e k = ( mg/ y) = (. 0 kg)(9. 8 m/s )/(. 0 5 m. ( 0 0 m)) = 39 N/m y = mg/ k = (3. 0 kg)(9. 8 m/s )/(39 N/m) y y = m y = y m =. 0 0 m m = m = 7. 5 cm e The length of the spring is 7.5 cm when a mass of 3.0 kg is attached to the spring. The position of the end of the spring is negative because it is below the origin, but length must be a positive number Model: Model the student (S) as a particle and the spring as obeying Hooke s law. Solve: According to Newton s second law the force on the student is S ( F ) = F F = ma spring on S G on S y spring on S G F = F + ma = mg + ma = (60 kg)(9. 8 m/s m/s ) = 768 N y Since Fspring on S = FS on spring = k y, k y = 768 N. This means y = (768 N)/(500 N/m) = 0. 3 m. y y 0.6. Model: Assume the spring is ideal and obeys Hooke s law. The stretch produced by hanging mass m is L L 0. Solve: For a hanging mass of m L = L + ( L L ) If we double the hanging mass to m, then and, indeed, for any mass nm, In particular for 3m 0 0 L = L + ( L L ) 0 0 L = L + nl ( L) n 0 0 L = L + 3( L L ) = L + 3L Assess: Even though one term is negative the answer won t be because L > L Model: Assume that the spring is ideal and obeys Hooke s law. We also model the 5.0 kg mass as a particle. We will use the subscript s for the scale and sp for the spring. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

11 Energy 0- Solve: (a) The scale reads the upward force F s on m that it applies to the mass. Newton s second law gives ( Fon m) y = Fs on m FG = 0 Fs on m = FG = mg = (5. 0 kg)(9. 8 m/s ) = 49 N (b) In this case, the force is ( F ) = F + F F = 0 0 N + k y mg = 0 on m y s on m sp G k = ( mg 0 N)/ y = (49 N 0 N)/0. 0 m = 450 N/m 3 The spring constant for the lower spring is N/m. (c) In this case, the force is ( F ) = F F = 0 k y mg = 0 on m y sp G y = mg/ k = (49 N)/(450 N/m) = m = 3. 4 cm Section 0.5 Elastic Potential Energy 0.8. Model: Assume an ideal spring that obeys Hooke s law. Solve: The elastic potential energy of a spring is defined as U s = k( s), where s is the magnitude of the stretching or compression relative to the unstretched or uncompressed length. Us = 0 when the spring is at its equilibrium length and s = 0. We have U = 00 J and k = 000 N/m. Solving for s: s s= U / k = (00 J)/000 N/m = m s 0.9. Model: Assume the spring is ideal and obeys Hooke s law. Then the potential energy of a stretched spring is Usp = k ( s ). Use ratios to solve this problem. Use primed variables for the new situation with the spring stretched three times as far. Solve: Usp k( s) k(3 s) = = = 3 = 9 U k( s) k( s) sp U sp = 9U = 9(. 0 J) = 8 J Assess: The stored energy scales with the square of the spring stretch. sp Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

12 0- Chapter Model: Assume an ideal spring that obeys Hooke s law. There is no friction, so the mechanical energy K + U s is conserved. Also model the book as a particle. The figure shows a before-and-after pictorial representation. The compressed spring will push on the book until the spring has returned to its equilibrium length. We put the origin of our coordinate system at the equilibrium position of the free end of the spring. The energy bar chart shows that the potential energy of the compressed spring is entirely transformed into the kinetic energy of the book. Solve: The conservation of energy equation K + Us = K + Us is mv + k( x xe) = mv + k( x xe) Using x = xe = 0 m and v = 0 m/s, this simplifies to kx. (50 N/m)(0 040 m) mv = k( x 0 m) v = = =. 0 m/s m ( kg) Assess: This problem cannot be solved using constant-acceleration kinematic equations. The acceleration is not a constant in this problem, since the spring force, given as Fs = k x, is directly proportional to x or x x. 0.. Model: Assume an ideal spring that obeys Hooke s law. Since there is no friction, the mechanical energy K + U s is conserved. Also, model the block as a particle. e Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

13 Energy 0-3 The figure shows a before-and-after pictorial representation. We have put the origin of our coordinate system at the equilibrium position of the free end of the spring. This gives us x= xe = 0 cm and x =. 0 cm. Solve: The conservation of energy equation K + Us = K + Us is mv + k( x xe) = mv + k( x xe) v = 0 m/s, x = x = 0 m, and x x = m, we get Using e e m k( x x ) mv x ( x x ) v k e = = e = That is, the compression is directly proportional to the velocity v. When the block collides with the spring with twice the earlier velocity ( v ), the compression will also be doubled to ( x xe) = (. 0 cm) = 4. 0 cm. Assess: This problem shows the power of using energy conservation over using Newton s laws in solving problems involving nonconstant acceleration. 0.. Model: Model the grocery cart as a particle and the spring as an ideal that obeys Hooke s law. We will also assume zero rolling friction during the compression of the spring, so that mechanical energy is conserved. The figure shows a before-and-after pictorial representation. The before situation is when the cart hits the spring in its equilibrium position. We put the origin of our coordinate system at this equilibrium position of the free end of the spring. This gives x= xe = 0 and ( x xe) = 60 cm. Solve: The conservation of energy equation K + Us = K + Us is mv + k( x xe) = mv + k( x xe) v = 0 m/s, ( x x ) = m, and x = x = 0 m gives: Using e e k 50 N/m k( x xe) = mv v= ( x xe) = (0. 60 m) = 3. 0 m/s m 0 kg 0.3. Model: Assume an ideal spring that obeys Hooke s law. There is no friction, and thus the mechanical ener- K + U + U is conserved. gy s g Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

14 0-4 Chapter 0 We place the origin of our coordinate system at the spring s compressed position y = 0. The rock leaves the spring with velocity v as the spring reaches its equilibrium position. Solve: (a) The conservation of mechanical energy equation is K + Us + Ug = K + Us + Ug mv + k( y ye) + mgy = mv + k( y ye) + mgy Using y = ye, y= 0 m, and v= 0 m/s, this simplifies to mv + 0J+ mgy =0 J + k( y ye) + 0 (0.400 kg) v + (0.400 kg)(9.8 m/s )(0.30 m) = (000 N/m)(0.30 m) v = 4.8 m/s (b) Let us use the conservation of mechanical energy equation once again to find the highest position ( y 3) of the rock where its speed ( v 3) is zero: K U K U mv mgy mv mgy 3+ g3 = + g = + v (4. 8 m/s) g( y y ) = v ( y y ) = = =. m g (9. 8 m/s ) If we assume the spring s length to be 0.5 m, then the distance between ground and fruit is. m+ 05. m= 7. m. This is much smaller than the distance of 5 m between fruit and ground. So, the rock does not reach the fruit, and the contestants go hungry Model: Model the jet plane as a particle, and the spring as an ideal that obeys Hooke s law. We will also assume zero rolling friction during the stretching of the spring, so that mechanical energy is conserved. The figure shows a before-and-after pictorial representation. The before situation occurs just as the jet plane lands on the aircraft carrier and the spring is in its equilibrium position. We put the origin of our coordinate system at the right free end of the spring. This gives x= xe = 0 m. Since the spring stretches 30 m to stop the plane, x xe = 30 m. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

15 Energy 0-5 Solve: The conservation of energy equation K + Us = K + Us for the spring-jet plane system is mv + k( x xe) = mv + k( x xe) v = 0 m/s, x = x = 0 m, and x x = 30 m yields Using e e k 60,000 N/m k( x xe) = mv v= ( x x) = (30 m) = 60 m/s m 5,000 kg Assess: A landing speed of 60 m/s or 0 mph is reasonable. Section 0.6 Energy Diagrams 0.5. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. The particle is released from rest at x =. 0 m. That is, K = 0 at x =. 0 m. Since the total energy is given by E = K + U, we can draw a horizontal total energy (TE) line through the point of intersection of the potential energy curve (PE) and the x =. 0 m line. The distance from the PE curve to the TE line is the particle s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: (a) We have E = 4. 0 J and this energy is a constant. For x< 0,. U > 40. J and, therefore, K must be negative to keep E the same (note that K = E U or K = 4. 0 J U). Since negative kinetic energy is unphysical, the particle cannot move to the left. That is, the particle will move to the right of x =. 0 m. (b) The expression for the kinetic energy is E U. This means the particle has maximum speed or maximum kinetic energy when U is minimum. This happens at x = 4. 0 m. Thus, (3. 0 J) 8. 0 J Kmax = E Umin = (40. J) (0. J) = 30. J mvmax = 30. J vmax = = = 73. m/s m kg The particle possesses this speed at x = 4. 0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at turning points of the motion Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. x =. 0 m and x = 6. 0 m. These are the Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

16 0-6 Chapter 0 The particle with a mass of 500 g is released from rest at A. That is, K = 0 at A. Since E = K + U = 0 J + U, we can draw a horizontal TE line through U = 5. 0 J. The distance from the PE curve to the TE line is the particle s kinetic energy. These values are transformed as the position changes, causing the particle to speed up or slow down, but the sum K + U does not change. Solve: The kinetic energy is given by E U, so we have ( )/ mv = E U v= E U m Using UB =. 0 J, UC = 3. 0 J, and UD = 0 J, we get vb = (50. J 0. J)/0500. kg= 35. m/s vc = (50. J 30. J)/0500. kg =. 8 m/s v = (5. 0 J 0 J)/ kg = 4. 5 m/s D 0.7. Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. For the speed of the particle at A that is needed to reach B to be a minimum, the particle s kinetic energy as it reaches the top must be zero. Similarly, the minimum speed at B for the particle to reach A obtains when the particle just makes it to the top with zero kinetic energy. Solve: (a) The energy equation KA + UA = Ktop + Utop is mva + UA = 0J+ Utop v = ( U U )/ m = (5.0 J.0 J)/0.00 kg = 7.7 m/s A top A (b) To go from point B to point A, KB + UB = Ktop + Utop is mvb + UB = 0 J + Utop v = ( U U )/ m = (5. 0 J 0 J)/0. 00 kg = 0. 0 m/s B top B Assess: The particle requires a higher kinetic energy to reach A from B than to reach B from A Model: For an energy diagram, the sum of the kinetic and potential energy is a constant. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

17 Energy 0-7 Since the particle oscillates between x =.0 mm and x = 8.0 mm, the speed of the particle is zero at these points. That is, for these values of x, E = U = 5. 0 J, which defines the total energy (TE) line. The distance from the potential energy (PE) curve to the TE line is the particle s kinetic energy. These values are transformed as the position changes, but the sum K + U does not change. Solve: The equation for total energy E = U + K means K = E U, so that K is maximum when U is minimum. We have Kmax = mvmax = 5. 0 J Umin v = (5. 0 J U )/ m = (5. 0 J. 0 J)/ kg = 63 m/s max Section 0.7 Elastic Collisions min 0.9. Model: We assume this is a one-dimensional collision that obeys the conservation laws of momentum and mechanical energy. Note that momentum conservation alone is not sufficient to solve this problem because the two final velocities ( v fx) and ( v fx) are unknowns and cannot be determined from one equation. Momentum conservation: m ( v ) + m ( v ) = m ( v ) + m ( v ) Solve: ix ix fx fx Energy conservation: m( vix) + m( vix) = m( vfx) + m( vfx) These two equations can be solved for ( v fx) and ( vfx ), as shown by Equations 0.39 through 0.43, to give m m 50 g 0 g ( vfx) = ( vix) = (. 0 m/s) = m/s m + m 50 g + 0 g m (50 g) ( v ) = ( ) = (. 0 m/s) =. 9 m/s + 50 g + 0 g fx vix m m Assess: These velocities are of a reasonable magnitude. Since both these velocities are positive, both balls move along the +x-direction Model: This is a case of a perfectly elastic collision between a proton and a carbon atom. The collision obeys the momentum as well as the energy conservation law. Solve: Momentum conservation: mp( vix) P + mc( vix) C = mp( vfx) P + mc( vfx) C Energy conservation: m ( v ) m ( v ) m ( v ) m ( v ) P ix P + C ix C = P fx P + C fx C Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

18 0-8 Chapter 0 These two equations can be solved, as described in the text through Equations 0.38 to 0.4: mp mc mp mp 7 7 ( vfx) P = mp + mc( vix) P = (. 0 0 m/s) = m/s mp + mc mp + mp mp mp 7 6 ( vfx) C = ( vix) P = (. 0 0 m/s) = 3. 0 m/s mp + mc mp + mp After the elastic collision the proton rebounds at m/s m/s and the carbon atom moves forward at 0.3. Model: In this case of a one-dimensional collision, the momentum conservation law is obeyed whether the collision is perfectly elastic or perfectly inelastic. Assume ball is initially moving right, in the positive direction. Solve: In the case of a perfectly elastic collision, the two velocities ( v fx) and ( vfx ) can be determined by combining the conservation equations of momentum and mechanical energy. By contrast, a perfectly inelastic collision involves only one final velocity v fx and can be determined from just the momentum conservation equation. (a) Momentum conservation: m( vix) + m( vix) = m( vfx) + m( vfx) Energy conservation: m( vix) + m( vix) = m( vfx) + m( vfx) These two equations can be solved as shown in Equations 0.38 through 0.4: m m (00 g) (300 g) ( v ) = ( ) = (0 m/s) =. 5 0 m/s + (00 g) + (300 g) fx vix m m m (00 g) ( v ) = ( ) = (0 m/s) = m/s + (00 g) + (300 g) fx vix m m (b) For the inelastic collision, both balls travel with the same final speed v fx. The momentum conservation equation p fx = pi x is ( m + m ) v = m ( v ) + m ( v ) fx ix ix 00 g vfx = (0 m/s) + 0 m/s =. 5 m/s 00 g g Assess: In the case of the perfectly elastic collision, the two balls bounce off each other with a speed of 5.0 m/s. In the case of the perfectly inelastic collision, the balls stick together and move together at.5 m/s Model: This is the case of a perfectly inelastic collision. Momentum is conserved because no external force acts on the system (clay + brick). We also represent our system as a particle. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

19 Energy 0-9 Solve: (a) The conservation of momentum equation pfx = pix is ( m + m ) v = m ( v ) + m ( v ) Using vix v0 vix ( ) = and ( ) = 0, we get fx ix ix m kg v v v v v fx = ( ix) = ( ix) = ( ix) = m+ m (. 0 kg kg) The brick is moving with speed v0. (b) The initial and final kinetic energies are given by K = m ( vx) + m ( vx) = ( kg) v + (. 0 kg)(0 m/s) = (0. 05 kg) v Kf = ( m+ m) vfx = (. 0 kg kg)( ) v0 = v0 K The percent of energy lost i Kf = 00% = 00% = 95% Ki Exercises and Problems i i i Model: We will take the system to be the person plus the earth. When a person drops from a certain height, the initial potential energy is transformed to kinetic energy. When the person hits the ground, if they land rigidly upright, we assume that all of this energy is transformed into elastic potential energy of the compressed leg bones. The maximum energy that can be absorbed by the leg bones is 00 J; this limits the maximum height. Solve: (a) The initial potential energy can be at most 00 J, so the height h of the jump is limited by mgh = 00 J For m = 60 kg, this limits the height to h = 00 J/mg = 00 J/(60 kg)(9. 8 m/s ) = m (b) If some of the energy is transformed to other forms than elastic energy in the bones, the initial height can be greater. If a person flexes her legs on landing, some energy is transformed to thermal energy. This allows for a greater initial height. Assess: There are other tissues in the body with elastic properties that will absorb energy as well, so this limit is quite conservative Model: Model your vehicle as a particle. Assume zero rolling friction, so that the sum of your kinetic and gravitational potential energy does not change as the vehicle coasts down the hill. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

20 0-0 Chapter 0 The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope is given, since these are not needed to solve the problem. All we need is the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that 35 km/hr = (35,000 m/3600 s) = 9. 7 m/s Solve: Using y f = 0 and the equation Ki + Ugi = Kf + Ugf we get mvi + mgyi = mvf + mgyf vi + gyi = vf f i i v = v + gy = (9. 7 m/s) + (9. 8 m/s )(5 m) = 9. 7 m/s = 7 km/h You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 9.7 m/s or 7 km/h at the bottom of the hill, when your speed at the top of the hill was 35 km/s, is reasonable. From the energy bar chart, we see that the initial potential energy is completely transformed into the final kinetic energy Model: This is case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the cannon ball falls. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

21 Energy 0- The figure shows a before-and-after pictorial representation. To express the gravitational potential energy, we put the origin of our coordinate system on the ground below the fortress. Solve: Using y f = 0 and the equation Ki + Ugi = Kf + Ugf we get mv mgy mv mgy v gy v i + i = f + f i + i = f f i i v = v + gy = (80 m/s) + (9. 8 m/s )(0 m) = 8 m/s Assess: Note that we did not need to use the tilt angle of the cannon, because kinetic energy is a scalar. Also note that using the energy conservation equation, we can find only the magnitude of the final velocity, not the direction of the velocity vector Model: Assume the spring is ideal and obeys Hooke s law. Then the potential energy of a stretched spring is Usp = k ( s ). The kinetic energy at the beginning is zero, and it is also zero at maximum height, so the spring potential energy at the beginning equals the gravitational potential energy at maximum height. We are given k = 950N/m. Solve: ( Us) i = ( U g) f k( s) = mgy k Putting this in y = mx + b form y = ( s) leads us to believe that a graph of y vs. ( s) would produce a mg straight line whose slope is k/mg and whose intercept is zero. The spreadsheet tells us the fit is very good and that the slope is m. k k 950 N/m m m kg 65 g mg =. = g( m ) = (9. 8 m/s )( m ) =. = Assess: 65 g seems reasonable, and we were happy to get a very small intercept on our best-fit line Model: For the ice cube sliding around the inside of a smooth pipe, the sum of the kinetic and gravitational potential energy does not change. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

22 0- Chapter 0 We use a coordinate system with the origin at the bottom of the pipe, that is, y = 0. The radius R of the pipe is 0 cm, and therefore ytop = y = R= 0.0 m. At an arbitrary angle θ, measured counterclockwise from the bottom of the circle, y = R Rcos θ. Solve: (a) The energy conservation equation K + Ug = K + Ug is mv + mgy = mv + mgy v = v + g( y y ) = (30. m/s) + (98. m/s )(0 m 00. m) = 5. m/s 3m/s. (b) Expressing the energy conservation equation as a function of θ : K( θ) + Ug ( θ) = K + Ug mv ( θ) + mgy( θ) = mv + 0 J v( θ) = v gy( θ) = v gr( cos θ) Using v = 3. 0 m/s, g = 9. 8 m/s, and R = 0. 0 m, we get v( θ) = 9. 96( cos θ) (m/s) Assess: Beginning with a speed of 3.0 m/s at the bottom, the marble s potential energy increases and kinetic energy decreases as it gets toward the top of the circle. At the top, its speed is.5 m/s. This is reasonable since some of the kinetic energy has been transformed into the marble s potential energy Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Solve: (a) The rubber band is stretched to the left since a positive spring force on the rock due to the rubber band results from a negative displacement of the rock. That is, ( Fsp) x = kx, where x is the rock s displacement from the equilibrium position due to the spring force F sp. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

23 Energy 0-3 (b) Since the F sp versus x graph is linear with a negative slope and can be expressed as Fsp = kx, the rubber band obeys Hooke s law. (c) From the graph, Fsp = 0 N for x = 0 cm. Thus, Fsp 0 N k = = = 00 N/m =.0 0 N/m x 0.0 m (d) The conservation of energy equation Kf + Usf = Ki + Usi for the rock is mv kx mv kx mv k m kx f + f = i + i f + (0 m) = (0 m/s) + i k 00 N/m vf = xi = (0. 30 m) = 9 m/s m kg Assess: Note that x i is x, which is the displacement relative to the equilibrium position, and x f is the equilibrium position of the rubber band, which is equal to zero Model: We will assume the knee extensor tendon behaves according to Hooke s Law and stretches in a straight line. Visiualize: The elastic energy stored in a spring is given by U s = k( s). Solve: For athletes, Us,athlete = k( s) = (33,000 N/m)(0. 04 m) = 7. 7 J For non-athletes, Us,non athlete = k( s) = (33,000 N/m)( m) = 8. 0 J The difference in energy stored between athletes and non-athletes is therefore 9.7 J. Assess: Notice the energy stored by athletes is over.5 times the energy stored by non-athletes Model: Model the block as a particle and the springs as ideal springs obeying Hooke s law. There is no friction, hence the mechanical energy K + Us is conserved. Note that xf = xe and xi xe = x. The before-and-after pictorial representations show that we put the origin of the coordinate system at the equilibrium position of the free end of the springs. Solve: The conservation of energy equation Kf + Usf = Ki + Usi for the single spring is mv k x x mv k x x f + ( f e) = i + ( i e) Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

24 0-4 Chapter 0 Using the value for v f given in the problem, we get mv0 + 0 J = 0 J + k( x) mv0 = k( x) Conservation of energy for the two-spring case: mvf + 0 J = 0 J + kx ( i xe) + kx ( i xe) f ( ) mv = k x Using the result of the single-spring case, mvf mv0 Vf v0 = = Assess: The block separates from the spring at the equilibrium position of the spring Model: Model the block as a particle and the springs as ideal springs obeying Hooke s law. There is no friction, hence the mechanical energy K + U is conserved. s The springs in both cases have the same compression. x We put the origin of the coordinate system at the equilibrium position of the free end of the spring for the single-spring case (a), and at the free end of the two connected springs for the two-spring case (b). Solve: The conservation of energy for the single-spring case: Kf + Usf = Ki + Usi mvf + k( xf xe) = mvi + k( xi xe) x = x = 0 m, v = 0 m/s, and v = v, this equation simplifies to Using f e i f 0 mv ( ) 0 = k x Conservation of energy in the case of the two springs in series, where each spring compresses by x/, is K U K U mv mv k x k x Using xf = xe = 0 m and vi = 0 m/s, we get f + sf = i + si f + 0 = i + ( /) + ( /) mv ( ) f = k x Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

25 Energy 0-5 Comparing the two results we see that V f = v 0 /. Assess: The block pushes on the spring until the spring has returned to its equilibrium length Model: Assume an ideal spring that obeys Hooke s law. There is no friction, and therefore the mechanical energy K + Us + Ug is conserved. The figure shows a before-and-after pictorial representation. We have chosen to place the origin of the coordinate system at the position where the ice cube has compressed the spring 0 cm. That is, y 0 = 0. Solve: (a) The energy conservation equation K + Us + Ug = K0 + Us0 + Ug0 is mv k x x mgy mv k x x mgy Using v = 0 m/s, y0 = 0 m, and v0 = 0 m/s, (b) Insert the values given + ( e e) + = 0 + ( e) + 0 k( x) mgy = k( x xe) y = = h mg k( x) (5 N/m)(0. 0 m) h = = = 5. 5 cm 6 cm mg ( kg)(9. 8 m/s ) Assess: The net effect of the launch is to transform the potential energy stored in the spring into gravitational potential energy. The block has kinetic energy as it comes off the spring, but we did not need to know this energy to solve the problem. The answer is independent of the angle of the slope Model: Model the two packages as particles. Momentum is conserved in both inelastic and elastic collisions. Kinetic energy is conserved only in a perfectly elastic collision. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

26 0-6 Chapter 0 Solve: For a package with mass m the conservation of energy equation is Using ( v0) m = 0 m/s and y= 0 m, K U K U m v mgy m v mgy + g = 0 + g0 ( ) m + = ( 0) m + 0 m ( v ) m = mgy0 ( v ) m = gy0 = (9. 8 m/s )(3. 0 m) = m/s (a) For the perfectly inelastic collision the conservation of momentum equation is Using ( v ) m = 0 m /s, we get p = p ( m+ m)( v ) = m( v ) + ( m)( v ) fx ix 3m m m ( v ) = ( v ) /3 =. 56 m/s 3m m The packages move off together at a speed of.6 m/s. (b) For the elastic collision, the mass m package rebounds with velocity m m ( v3) m = ( v) m = (7.668 m/s) =.56 m/s m+ m 3 The negative sign with ( v 3) m shows that the package with mass m rebounds and goes to the position y 4. We can determine y 4 by applying the conservation of energy equation as follows. For a package of mass m: K U K U m v mgy m v mgy f + gf = i + gi ( 4) m + 4 = ( 3) m + 3 Using ( v3) m =. 55 m/s, y3 = 0 m, and ( v4) m = 0 m/s, we get mgy4 = m (. 56 m/s) y4 = 33 cm Model: Model the marble and the steel ball as particles. We will assume an elastic collision between the marble and the ball, and apply the conservation of momentum and the conservation of energy equations. We will also assume zero rolling friction between the marble and the incline. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

27 Energy 0-7 This is a two-part problem. In the first part, we will apply the conservation of energy equation to find the marble s speed as it exits onto a horizontal surface. We have put the origin of our coordinate system on the horizontal surface just where the marble exits the incline. In the second part, we will consider the elastic collision between the marble and the steel ball. Solve: The conservation of energy equation K + Ug = K0 + Ug0 gives us: mm( v) M + m Mgy= mm( v0) M + mmgy0 Using ( v0) M= 0 m/s and y= 0 m, we get ( v) M= gy0 ( v) M= gy0. When the marble collides with the steel ball, the elastic collision gives the ball velocity mm mm ( v) S= ( v) M= gy0 mm + ms mm + ms Solving for y 0 gives mm + ms y0= ( v) S = 0.58 m = 5.8 cm g mm Model: Assume an ideal spring that obeys Hooke s law. Since this is a free-fall problem, the mechanical energy K + Ug + Us is conserved. Also, model the safe as a particle. We have chosen to place the origin of our coordinate system at the free end of the spring, which is neither stretched nor compressed. The safe gains kinetic energy as it falls. The energy is then converted into elastic potential energy as the safe compresses the spring. The only two forces are gravity and the spring force, which are both conservative, so energy is conserved throughout the process. This means that the initial energy as the safe is released equals the final energy when the safe is at rest and the spring is fully compressed. Solve: The conservation of energy equation K + Ug + Us = K0 + Ug0 + Us0 is mv + mg( y y e) + k( y ye) = mv0 + mg( y0 ye) + k( ye ye) Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

28 0-8 Chapter 0 Using v= v0 = 0 m/s and y e = 0 m, the above equation simplifies to mgy ky mgy 0 y ( m) + = 0 mg( y ) (000 kg)(9. 8 m/s )(. 0 m. ( 0 50 m)) 5 5 k = = = N/m. 0 0 N/m y Assess: By equating energy at these two points, we do not need to find how fast the safe was moving when it hit the spring Model: Assume an ideal spring that obeys Hooke s law. There is no friction and hence the mechanical ener- ( v ) =. m/s +. 0 m/s =. 0 m/s is conserved. gy fx Solve: (a) When releasing the block suddenly, K + Us + Ug = K + Us + Ug mv + k( y y e) + mgy = mv + k( y ye) + mgy Using v = 0 m/s, v= 0 m/s, and y = y e, we get 0 J + (490 N/m)( y y) + mgy = 0 J + 0 J + mgy (45 N/m)( y y) = mg( y y) (45 N/m)( y y ) = (5. 0 kg)(9. 8 m/s )( y y ) ( y y ) = 0. 0 m (b) When lowering the block gently until it rests on the spring, the block reaches a point of static equilibrium. mg (5. 0 kg)(9. 8 m/s ) Fnet = k y mg = 0 y = = = 0. 0 m k 490 N/m (c) In part (b), at a point 0.0 m down, the forces balance. But in part (a) the block has kinetic energy as it reaches 0.0 m. So the block continues on past the equilibrium point until all the gravitational potential energy is stored in the spring Model: Assume an ideal spring that obeys Hooke s law. Also assume zero rolling friction between the roller coaster and the track, and a particle model for the roller coaster. Since no friction is involved, the mechanical energy K + U + U is conserved. s g We have chosen to place the origin of the coordinate system on the end of the spring that is compressed and touches the roller coaster car. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

29 Energy 0-9 Solve: (a) The energy conservation equation for the car going to the top of the hill is K + Ug + Us = K0 + Ug0 + Us0 mv + mgy + k( xe xe) = mv0 + mgy0 + k( x0 xe) Noting that y0 = 0 m, v = 0 m/s, v= 0 m/s, and x0 xe =. 0 m, we obtain 0 J + mgy + 0 J = 0 J + 0 J + k(. 0 m). mgy (400 kg)(9 8 m/s )(0 m) 4 k = = = N/m (. 0 m) (. 0 m) 4 4 We now increase this value for k by 0% for safety, giving a value of N/m. 0 N/m. (b) The energy conservation equation K3 + Ug3 + Us3 = K0 + Ug0 + Us0 is mv mgy k x x mv mgy k x x y =. 5 0 m, v = 0 m/s, y = 0 m, and x x =. 0 m, we get Using e ( e e) = ( 0 e) mv3 + mg(. 5 0 m) + 0 J = 0 J + 0 J + k( x0 xe) (400 kg) v (400 kg)(9 8 m/s )(5 0 m) ( 56 0 N/m)( 0 m) v = 7. 7 m/s 8 m/s = Model: Since there is no friction, the sum of the kinetic and gravitational potential energy does not change. Model Julie as a particle. We place the coordinate system at the bottom of the ramp directly below Julie s starting position. From geometry, Julie launches off the end of the ramp at a 30º angle. Solve: Energy conservation: K + Ug = K0 + Ug0 mv + mgy = mv0 + mgy0 Using v0 = 0 m/s, y0 = 5 m, and y= 3 m, the above equation simplifies to mv mgy mgy0 v g( y0 y) (9 8 m/s )(5 m 3 m) 0 77 m/s + = = =. =. We can now use kinematic equations to find the touchdown point from the base of the ramp. First we ll consider the vertical motion: y = y+ vy( t t) + ay( t t) 0 m = 3 m + ( vsin30 )( t t) + ( 9.8 m/s )( t t) ( 0.77 m/s) sin30 (3 m) ( t t) ( t t) = 0 (4.9 m/s ) (4.9 m/s ) Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

30 0-30 Chapter 0 ( t t ) (.9 s)( t t ) (0.6 s ) =0 ( t t ) =.377 s For the horizontal motion: x = x+ vx( t t) + ax( t t) x x = ( vcos30 )( t t ) + 0 m = (0. 77 m/s)(cos30 )(. 377 s) = 43 m Assess: Note that we did not have to make use of the information about the circular arc at the bottom that carries Julie through a 90 turn Model: We assume the spring to be ideal and to obey Hooke s law. We also treat the block (B) and the ball (b) as particles. In the case of an elastic collision, both the momentum and kinetic energy equations apply. On the other hand, for a perfectly inelastic collision only the equation of momentum conservation is valid. Place the origin of the coordinate system on the block that is attached to one end of the spring. The before-and-after pictorial representations of the elastic and perfectly inelastic collision are shown in figures (a) and (b), respectively. Solve: (a) For an elastic collision, the ball s rebound velocity is m m 80 g ( v ) = ( ) = (5. 0 m/s) = m/s 0 g b B f b vi b mb + mb The ball s speed is 3.3 m/s. (b) An elastic collision gives the block speed m 40 g ( v ) = ( ) = (5 0 m/s) 667 m/s 0 g. =. B f B vi b mb + mb To find the maximum compression of the spring, we use the conservation equation of mechanical energy for the block + spring system. That is K + Us = K0 + Us0: mb( vf) B + kx ( x 0) = mb( vf) B + kx ( 0 x0) 0 + kx ( x0) = mb( vf) B + 0 ( x x ) = (0. 00 kg)(. 667 m/s) /(0 N/m) =. 8 cm 0 Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.

31 Energy 0-3 (c) Momentum conservation pf = pi for the perfectly inelastic collision means ( m + m ) v = m ( v ) + m ( v ) B B f b i b B i B (0. 00 kg kg) v = (0. 00 kg)(5. 0 m/s) + 0 m/v v = m/s f The maximum compression in this case can now be obtained using the conservation of energy equation K + U s = K 0 + U s0 : 0 J + (/) k( x) = (/)( m + m ) v + 0 J B b f mb + mb 0. 0 kg x= v f = ( m/s) = m = 6. 5 cm k 0 N/m Model: Assume an ideal spring that obeys Hooke s law. We treat the bullet and the block in the particle model. For a perfectly inelastic collision, the momentum is conserved. Furthermore, since there is no friction, the mechanical energy of the system (bullet + block + spring) is conserved. f We place the origin of our coordinate system at the end of the spring that is not anchored to the wall. Solve: (a) Momentum conservation for perfectly inelastic collision states pf = pi. This means m ( m+ M) vf= m( vi) m+ M( vi) M ( m+ M) vf= mvb+ 0 kg m/s vf= vb m+ M where we have used v B for the initial speed of the bullet. The mechanical energy conservation equation K + Us = Ke + Use as the bullet embedded block compresses the spring is: mv ( f ) + kx ( xe) = ( m+ M)( vf) + kx ( e xe) m m + M kd B B ( ) 0 J + kd = ( m + M ) v + 0 J v = m+ M m (b) Using the above formula with m= 5. 0 g, M =. 0 kg, k = 50 N/m, and d = 0 cm, v B = ( kg +. 0 kg)(50 N/m)(0. 0 m) /( ) =. 0 0 m/s (c) The fraction of energy lost is mv B ( m + M ) vf m+ M vf m+ M m = = mv m vb m m+ M B m kg = = = m+ M ( kg +. 0 kg) That is, during the perfectly inelastic collision 99.75% of the bullet s energy is lost. The energy is dissipated inside the block. Although it is common to say, The energy is lost to heat, in the next chapter we ll see that it is more accurate to say, The energy is transformed to thermal energy. Copyright 03 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.