Chapter 10, Solutions, Physics 121

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1 Chapter 0, Solutions, Physics 0.. Model: We will use the particle model for the bullet (B) and the bowling ball (BB). Solve: For the bullet, For the bowling ball, K = m v = (0.0 kg)(500 m/s) = 50 J B B B KBB = mbbvbb = (0 kg)(0 m/s) = 500 J 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 th dependence. Although the mass of the bullet 000 times smaller than the mass of the bowling ball, its speed 50 times larger.

2 0.6. Model: Th a case of free fall, so the sum of the kinetic and gravitational potential energy does not change as the ball res and falls. 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 + U g the same during free fall: Kf + Ugf = Ki + Ugi. We have (a) mv + mgy = mv + mgy 0 0 ( ) y = v v g = [(0 m/s) (0 m/s) ]/( 9.8 m /s ) = 5.0 m 0 5. m therefore the maximum height of the ball above the window. Th 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) v 3 = (0 m/s) + (9.8 m /s )[0 m ( 0 m)] = 49 m /s Th means the magnitude of v 3 equal to. m/s. Assess: Note that the ball s speed as it passes the window on its way down the same as the speed with which it was tossed up, but in the opposite direction Model: Model the ball as a particle undergoing rolling motion with zero rolling friction. The sum of the ball s kinetic and gravitational potential energy, therefore, does not change during the rolling motion. Solve: Since the quantity K + U g does not change during rolling motion, the energy conservation equations apply. For the linear segment the energy conservation equation K0 + Ug0 = K + Ug mv mgy + = mv mgy For the parabolic part of the track, K + Ug = K + Ug + = + = m v mg(0 m) m(0 m/s) mgy0 mv mgy0 mv + mgy = mv + mgy mv + mg(0 m) = m(0 m /s) + mgy mv = mgy Since from the linear segment we have m v mgy0, we get mgy 0 mgy or y y0.0 m. = = = = Thus, the ball rolls up to exactly the same height as it started from. Assess: Note that th result independent of the shape of the path followed by the ball, provided there no rolling friction. Th result an important consequence of energy conservation.

3 0.5. Model: Assume that the spring 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. Solve: (a) The scale reads the upward force F s on m that it applies to the mass. Newton s second law gives (b) In th case, the force (c) In th case, the force ( Fon m) = Fs on m w = 0 Fs on m = w = mg = (5.0 kg)(9.8 m/s ) = 49 N y ( F ) = F + F w = 0 0 N + k y mg = 0 on m y s on m sp k = ( mg 0 N)/ y = (49 N 0 N)/0.0 m = 450 N/m ( F ) = F w = 0 k y mg = 0 on m y sp y = mg/ k = (49 N)/(450 N/ m) = m = 3.4 cm 0.9. Model: For an energy diagram, the sum of the kinetic and potential energy a constant. The particle released from rest at x =.0 m. That, K = 0 at x =.0 m. Since the total energy 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 dtance from the PE curve to the TE line 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 = 5.0 J and th energy a constant. For x <.0, U > 5.0 J and, therefore, K must be negative to keep E the same (note that K = E U or K = 5.0 J U). Since negative kinetic energy unphysical, the particle can not move to the left. That, the particle will move to the right of x =.0 m. (b) The expression for the kinetic energy E U. Th means the particle has maximum speed or maximum kinetic energy when U minimum. Th happens at x =.0 m. Thus, (4.0 J) 8.0 J Kmax = E Umin = (5.0 J) (.0 J) = 4.0 J mvmax = 4.0 J vmax = = = 0 m/s m 0.00 kg The particle possesses th speed at x =.0 m. (c) The total energy (TE) line intersects the potential energy (PE) curve at x =.0 m and x = 6.0 m. These are the turning points of the motion.

4 0.33. 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. The figure shows a before-and-after pictorial representation. Note that neither the shape of the hill nor the angle of the downward slope given, since these are not needed to solve the problem. All we need the change in potential energy as you and your vehicle descend to the bottom of the hill. Also note that 35 km / hr = (35000 m / 3600 s) = 9.7 m /s Solve: Using y f = 0 and the equation Ki + Ugi = Kf + Ugf we get mv + mgy = mv + mgy v + gy = v f i i i i f f i i f v = v + gy = (9.7 m/s) + (9.8 m/s)(5 m) = 9.7 m/s = 7 km/ hr You are driving over the speed limit. Yes, you will get a ticket. Assess: A speed of 9.7 m/s or 7 km/hr at the bottom of the hill, when your speed at the top of the hill was 35 km/s, reasonable. From the energy bar chart, we see that the initial potential energy completely transformed into the final kinetic energy.

5 0.37. Model: Assume that the rubber band behaves similar to a spring. Also, model the rock as a particle. Please refer to Figure P0.37. Solve: (a) The rubber band stretched to the left since a positive spring force on the rock due to the rubber band results from a negative dplacement of the rock. That, (F sp ) x = kx, where x the rock s dplacement from the equilibrium position due to the spring force F sp. (b) Since the F sp versus x graph linear with a negative slope and can be expressed as F sp = kx, the rubber band obeys Hooke s law. (c) From the graph, F sp = 0 N for x = 0 cm. Thus, k F 0 N x 0.0 m sp = = = 00 N/m (d) The conservation of energy equation K f + U sf = K i + U si for the rock mv + kx = mv + kx mv + k(0 m) = m(0 m/s) + kx f f i i f i k 00 N/m vf = xi = (0.30 m) = 9.0 m/s m 0.05 kg Assess: Note that x i x, which the dplacement relative to the equilibrium position, and x f the equilibrium position of the rubber band, which equal to zero.

6 0.39. Model: Model the block as a particle and the springs as ideal springs obeying Hooke s law. There no friction, hence the mechanical energy K + U s conserved. 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: K + U = K + U mv + k( x x ) = mv + k( x x ) f sf i si f f e i i e Using x f = x e = 0 m, v i = 0 m/s, and v f = v 0, th equation simplifies to mv = k ( x ) 0 Conservation of energy in the case of the two springs in series, where each spring compresses by x/, Using x f = x e = 0 m and v i = 0 m/s, we get K + U = K + U mv + 0 = mv + k( x/) + k( x/) f sf i si f i mv ( ) f = k x 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.

7 0.50. Model: Since there 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 + v ( t t ) + a ( t t ) 0 m = 3 m + ( v sin30 )( 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 ) For the horizontal motion: y y ( t t ) (.9 s)( t t ) (0.6 s ) = 0 ( t t ) =.377 s x = x + v ( t t ) + a ( t t ) x x x x = ( v cos30 )( t t ) + 0 m = (0.77 m/s)(cos30 )(.377 s) = 4.7 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.

8 0.53. Model: Because the track frictionless, the sum of the kinetic and gravitational potential energy does not change during the car s motion. We place the origin of the coordinate system at the ground level directly below the car s starting position. Th a two-part problem. If we first find the maximum speed at the top of the hill (point C), we can use energy conservation to find the maximum initial height. Solve: Because the its motion circular, at the top of the hill the car has a downward-pointing centripetal acceleration a = ( mv / r) ˆj. Newton s second law at the top of the hill r c mv mv v ( Fnet ) y = ny + wy = n mg = m( ac) y = n = mg = m g R R R If v = 0 m /s, n = mg as expected in static equilibrium. As v increases, n gets smaller the apparent weight of the car and passengers decreases as they go over the top. But n has to remain positive for the car to be on the track, so the maximum speed v max occurs when n 0. We see that vmax = gr. Now we can use energy conservation to relate point C to the starting height: KC + UC = Ki + Ui mvc + mgyc = mv + mgyi mgr + mgr = 0 J + mghmax, 3 where we used v C = v max and y C = R. Solving for h max gives h = R (b) If R = 0 m, then h max = 5 m. max.

9 0.54. Model: Th a two-part problem. In the first part, we will find the critical velocity for the ball to go over the top of the peg without the string going slack. Using the energy conservation equation, we will then obtain the gravitational potential energy that gets transformed into the critical kinetic energy of the ball, thus determining the angle θ. We place the origin of our coordinate system on the peg. Th choice will provide a reference to measure gravitational potential energy. For θ to be minimum, the ball will just go over the top of the peg. Solve: The two forces in the free-body force diagram provide the centripetal acceleration at the top of the circle. Newton s second law at th point mv w + T = r where T the tension in the string. The critical speed v c at which the string goes slack found when T 0. In th case, The ball should have kinetic energy at least equal to mvc mg = v = gr = gl r C 3 L mvc = mg 3 for the ball to go over the top of the peg. We will now use the conservation of mechanical energy equation to get the minimum angle θ. The equation for the conservation of energy f c f 3 i c K + U = K + U mv + mgy = mv + mgy f gf i gi f f i i Using v = v, y = L, v = 0, and the above value for v, we get mg L + mg L = mgy L i yi = 3 3 That, the ball a vertical dtance L above the peg s location or a dtance of L L L = 3 6 below the point of suspension of the pendulum, as shown in the figure on the right. Thus, L/6 cosθ = = θ = 80.4 L 6

10 0.56. Model: Model La (L) and the bobsled (B) as particles. We will assume the ramp to be frictionless, so that the mechanical energy of the system (La + bobsled + spring) conserved. Furthermore, during the collion, as La leaps onto the bobsled, the momentum of the La + bobsled system conserved. We will also assume the spring to be an ideal one that obeys Hooke s law. We place the origin of our coordinate system directly below the bobsled s initial position. Solve: (a) Momentum conservation in La s collion with bobsled states p = p0, or ( m + m ) v = m ( v ) + m ( v ) ( m + m ) v = m ( v ) + 0 L B L 0 L B 0 B L B L 0 L m L 40 kg v = ( v0 ) L = ( m/s) = 8.0 m/s ml + mb 40 kg + 0 kg The energy conservation equation: K + Us + Ug = K + Us + Ug ( m + m ) v + k( x x ) + ( m + m ) gy = ( m + m ) v + k( x x ) + ( m + m ) gy L B e L B L B e e L B Using v = 0 m/s, k = 000 N/m, y = 0 m, y = (50 m) sin 0 = 7. m, v = 8.0 m/s, and (m L + m B ) = 60 kg, we get 0 J + (000 N / m)( x xe) + 0 J = (60 kg)(8.0 m /s) + 0 J + (60 kg)(9.8 m /s )(7. m) Solving th equation yields ( x xe) = 3.46 m. (b) As long as the ice slippery enough to be considered frictionless, we know from conservation of mechanical energy that the speed at the bottom depends only on the vertical descent y. Only the ramp s height h important, not its shape or angle Model: We will use the conservation of mechanical energy. The potential energy (U) of the nitrogen atom as a function of z exhibits a double-minimum behavior; the two minima correspond to the nitrogen atom s position on both sides of the plane containing the three hydrogens. Solve: (a) At room temperature, the total energy line below the hill in the center of the potential energy curve. That, the nitrogen atom does not have sufficient energy to pass from one side of the molecule to the other. There s a stable equilibrium position on either side of the hydrogen-atom plane at the points where U = 0. Since E > 0, the nitrogen atom will be on one side of the plane and will make small vibrations back and forth along the z-ax that, toward and away from the hydrogen-atom plane. In the figure above, the atom oscillates between points A and B. (b) The total energy line now well above the hill, and the turning points of the nitrogen atom s motion (where the total energy line crosses the potential curve) are at points C and D. In other words, the atom oscillates from one side of the H 3 plane to the other. It slows down a little as it passes through the plane of hydrogen atoms, but it has sufficient energy to get through.

11 0.63. A.5 kg ball thrown upward at a speed of 4.0 m/s from a height of 8 cm above a vertical spring. When the ball comes down it lands on and compresses the spring. If the spring has a spring constant of k = 600 N/m, by how much it compressed? Model: Model the sled as a particle. Because there no friction, the sum of the kinetic and gravitational potential energy conserved during motion. Place the origin of the coordinate system at the center of the hemphere. Then y 0 = R and, from geometry, y = Rcos φ. Solve: The energy conservation equation K + U = K0 + U0 mv + mgy = mv0 + mgy0 mv + mgr cosφ = mgr v = gr( cos φ) (b) If the sled on the hill, it moving in a circle and the r-component of magnitude net /. F r net has to point to the center with F = mv R Eventually the speed gets so large that there not enough force to keep it in a circular trajectory, and that the point where it flies off the hill. Consider the sled at angle φ. Establh an r-ax pointing toward the center of the circle, as we usually do for circular motion problems. Newton s second law along th ax requires: mv ( Fnet ) r = wcosφ n = mgcosφ n = mar = R mv v n = mgcosφ = m gcosφ R R The normal force decreases as v increases. But n can t be negative, so the fastest speed at which the sled stays on the hill the speed v max that makes n 0. We can see that vmax = gr cos φ. (c) We now know the sled s speed at angle φ, and we know the maximum speed it can have while remaining on the hill. The angle at which v reaches v max the angle φ max at which the sled will fly off the hill. Combining the two expressions for v and v max gives: gr( cos φ) = grcosφ R( cos φ ) = Rcosφ max cosφmax = φmax = cos = max