PH211 Chapter 10 Solutions

Transcription

1 PH Chapter 0 Solutions 0.. Model: We will use the particle model for the bullet (B) and the running student (S). Solve: For the bullet, K B = m v = B B (0.00 kg)(500 m/s) = 50 J For the running student, K S = m v = S S (75 kg)(5.5 m/s) = 06 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 7500 times smaller than the mass of the student, its speed more than 90 times larger 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: K f + U gf = K i + U gi. We have (a) = 0 0 y = v ( 0 v ) g = [(0 m/s) (0 m/s) ]/( 9.8 m / s ) = 5.0 m 5. m therefore the maximum height of the ball above the window. Th 5. m above the ground.

2 (b) = 0 0 Since y = y 0 = 0, we get for the magnitudes v = v 0 = 0 m / s. (c) 3 3 = 0 0 v 3 + gy 3 = v 0 + gy 0 v 3 = v 0 + g( y 0 y 3 ) 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 puck as a particle. Since the ramp frictionless, the sum of the puck s kinetic and gravitational potential energy does not change during its sliding motion. Solve: The quantity K + U g the same at the top of the ramp as it was at the bottom. The energy conservation equation K f + U gf = K i + U gi f f = i i v i = v f + g( y f y i ) v i = (0 m/s) + (9.8 m/s )(.03 m 0 m) = 0. m /s v i = 4.5 m/s Assess: An initial push with a speed of 4.5 m/s 0 mph to cover a dtance of 3.0 m up a 0 ramp seems reasonable.

3 0.8. 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 (F on S ) y = F spring on S F G = ma y F spring on S = F G + ma y = mg + ma y = (60 kg)(9.8 m/s m/s ) = 768 N Since F spring on S = F S on spring = kδy, kδy = 768 N. Th means Δy = (768 N)/(500 N/m) = 0.3 m. 0.. Model: Assume an ideal spring that obeys Hooke s law. Since there no friction, the mechanical energy K + U s conserved. Also, model the block as a particle. 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. Th gives us x = x e = 0 cm and x =.0 cm. Solve: The conservation of energy equation K + U s = K + U s mv + k(x x e ) = mv + k(x x e ) Using v = 0 m/s, x = x e = 0 m, and x x e = 0.00 m, we get k(x x e ) = mv Δx = (x x e ) = m k v

4 That, the compression 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 x e ) = (.0 cm) = 4.0 cm. Assess: Th problem shows the power of using energy conservation over using Newton s laws in solving problems involving nonconstant acceleration Model: Th the case of a perfectly inelastic collion. Momentum conserved because no external force acts on the system (clay + brick). We also represent our system as a particle. Solve: (a) The conservation of momentum equation p fx = p ix Using (v ix ) = v 0 and (v ix ) = 0, we get v fx = The brick moving with speed 0.048v 0. (m + m )v fx = m (v ix ) + m (v ix ) m kg (v m + m ix ) = (.0 kg kg) (v ) = (v ) = v ix ix 0 (b) The initial and final kinetic energies are given by K i = m (v ) + ix m (v ) = ix (0.050 kg)v + (.0 kg) 0 ( 0 m/s) = (0.05 kg)v 0 K f = (m + m )v = fx (.0 kg kg)(0.0476) v 0 = v 0 The percent of energy lost = K i K f K i % = % = 95% 0.3. Model: For an energy diagram, the sum of the kinetic and potential energy a constant. Since the particle oscillates between x =.0 mm and x = 8.0 mm, the speed of the particle zero at these points. That, for these values of x, E = U = 5.0 J, which defines the total energy (TE) line. The dtance from the potential energy (PE) curve to the TE line 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 maximum when U minimum. We have

5 K max = mv = 5.0 J U max min v max = (5.0 J U min )/m = (5.0 J.0 J)/0.000 kg = 63 m/s 0.4. Model: Model the two packages as particles. Momentum conserved in both inelastic and elastic collions. Kinetic energy conserved only in a perfectly elastic collion. Solve: For a package with mass m the conservation of energy equation Using (v 0 = 0 m / s and y = 0 m, K + U g = K 0 + U g0 m(v ) + mgy m = m(v ) + mgy 0 m 0 m(v ) = mgy m 0 (v = gy 0 = (9.8 m/s )(3.0 m) = m/s (a) For the perfectly inelastic collion the conservation of momentum equation Using (v ) m = 0 m / s, we get p fx = p ix (m + m)(v ) 3m = m(v + (m)(v ) m (v ) 3m = (v / 3 =.56 m/s The packages move off together at a speed of.6 m/s. (b) For the elastic collion, the mass m package rebounds with velocity (v 3 = m m m + m (v ) = m/s m 3 ( ) =.56 m/s The negative sign with (v 3 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 f + U gf = K i + U gi m(v ) + mgy 4 m 4 = m(v ) + mgy 3 m 3 Using (v 3 =.55 m/s, y 3 = 0 m, and (v 4 = 0 m/s, we get

6 mgy 4 = m(.56 m/s) y 4 = 33 cm 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 + U g = K 0 + U g0 = 0 0 Using v 0 = 0 m/s, y 0 = 5 m, and y = 3 m, the above equation simplifies to = mgy 0 v = g( y 0 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 y (t t ) + a y (t t ) 0 m = 3 m + (v sin30 )(t t ) + ( 9.8 m / s )(t t ) For the horizontal motion: (0.77 m/s)sin30 (3 m) (t t ) (t (4.9 m/s ) t ) (4.9 m/s ) = 0 (t t ) (.9 s)(t t ) (0.6 s ) = 0 (t t ) =.377 s x = x + v x (t t ) + a x (t t ) x x = (v cos30 )(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: 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.

7 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 = p 0, or (m L )v = m L (v 0 ) L (v 0 ) B (m L )v = m L (v 0 ) L + 0 m v = L m L + m B (v ) = 40 kg 0 L 40 kg + 0 kg ( m/s) = 8.0 m/s The energy conservation equation: K + U s + U g = K + U s + U g (m + m )v + L B k(x x e ) + (m L )gy = (m + m )v + L B k(x x e e ) + (m L )gy Using v = 0 m/s, k = 000 N/m, y = 0 m, y = (50 m)sin0 = 7. m, v = 8.0 m/s, and (m L m B ) = 60 kg, we get 0 J + (000 N / m)(x x e ) + 0 J = (60 kg)(8.0 m / s) + 0 J + (60 kg)(9.8 m / s )(7. m) Solving th equation yields (x x e ) = 3.5 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 Δy. Only the ramp s height h important, not its shape or angle 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.

8 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 F G + T = mv 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 mg = mv C v r C = gr = gl 3 mv = C mg L 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 K f + U gf = K i + U gi f f = i i Using v f = v c, y f = 3 L, v i = 0, and the above value for v C, we get mg L 3 + mg L 3 = mgy i y i = L That, the ball a vertical dtance L above the peg s location or a dtance of L 3 L = L 6 below the point of suspension of the pendulum, as shown in the figure on the right. Thus, cosθ = L/6 L = 6 θ = Model: Assume an ideal spring that obeys Hooke s law. There no friction, hence the mechanical energy K + U g + U s conserved. We have chosen to place the origin of the coordinate system at the point of maximum compression. We will use lengths along the ramp with the variable s rather than x.

9 Solve: (a) The conservation of energy equation K + U g + U s = K + U g + U s + k(δs) = + k(0 m) m(0 m/s) + mg(0 m) + k(δs) = m(0 m/s) + mg(4.0 m + Δs)sin J Th gives the quadratic equation: (50 N/m)(Δs) = (0 kg)(9.8 m/s )(4.0 m + Δs) (5 N/m)(Δs) (49 kg m/s )Δs 96 kg m /s = 0 Δs =.46 m and.07 m (unphysical) The maximum compression.46 m. (b) We will now apply the conservation of mechanical energy to a point where the vertical position y and the block s velocity v. We place the origin of our coordinate system on the free end of the spring when the spring neither compressed nor stretched. mv + mgy + k(δs) = + k(0 m) mv + mg( Δs sin30 ) + k(δs) = 0 J + mg(4.0 m sin30 ) + 0 J k(δs) (mg sin30 )Δs + mv mg sin30 (4.0 m) = 0 To find the compression where v maximum, take the derivative of th equation with respect to Δs: Since dv = 0 at the maximum, we have dδs k (Δs) (mg sin30 ) + dv m v dδs 0 = 0 Δs = (mg sin30 )/k = (0 kg)(9.8 m/s )(0.5)/(50 N/m) = 9.6 cm