# Problem Set #8 Solutions

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2 8.01L Problem Set #8 Solutions, p. 2 speed v 0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is µ r. What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods. Let the distance on the truck ramp be d. Also, let s define the point A as the starting point at which the truck is a distance L from the truck ramp and its velocity is v 0, and B the point at which the truck stops. We use energy transformation law on the truck+earth system: Now: E = W non cons (1) E B E A = f k r (2) E B E A = µ k Nd where N = mg cos β (3) E A = K A + U A = 1 2 mv2 0 + mgl sin α (4) E B = K B + U B = mgd sin β (5) since the velocity at point B is zero. Using equation 3 So, from equations 4 and 5: E B E A = µ k mgd cos β (6) mgd sin β mgl sin α 1 2 mv2 0 = µ k mgd cos β (7) gd(sin β + µ cos β) = gl sin α v2 0 (8) Note: as expected, d = L, if d = L sin α v2 0/g sin β + µ cos β α = β, µ = 0 v 0 = 0. (9) (10)

3 8.01L Problem Set #8 Solutions, p : Loop A small block with mass m slides in a vertical circle of radius R on the inside of a circular vertical track. There is no friction between the track and the block. At the bottom of the block s path, the normal force the track exerts on the block has magnitude N b. What is the magnitude of the normal force (call it N t ) that the track exerts on the block when it is at the top of its path? A block is on the inside of a circular track, moving at speed v. Let the speed at the bottom be v b and that of the top be v t We want expressions for the normal force at top & bottom. Use Newton s 2nd law. Top: Bottom: Fy = mg N t = mv2 t R N t = mv2 t R (11) mg (12) Fy = N b mg = mv2 b R N b = mg + mv2 b (14) R Now we need to solve for the velocities. Since we are given N b and m, we can use equation 14 to solve for v b : (13) v 2 b = R m (N b mg) (15) Now to find the solutions to the normal force at the top, we need to find v t. Using conservation of energy between the top and the bottom: KE b + U b = KE t + U t (16) 1 2 mv2 b + 0 = 1 2 mv2 t + 2mgR (17) vt 2 = vb 2 4gR (18)

4 8.01L Problem Set #8 Solutions, p. 4 Now finally we find N t, using equation 14 one more time: N t = mv2 b R 5 mg = N b 6mg. (19) 8-3: Pendulum Young and Freedman (13th edition) problem 7.82, page 239. (same as Young and Freedman (12th edition) problem 7.14, page 240.) Note: Young and Freedman stated this problem with numerical values m = 0.12 kg, l = 0.80 m, and θ = 45. You, however, are being asked to solve the problem symbolically. A small rock with mass m is fastened to a massless string with length l to form a pendulum. The pendulum is swinging so as to make a maximum angle of θ with the vertical. Air resistance is negligible. (a) What is the speed of the rock when the string passes through the vertical position? Choose y = 0 at the lowest point of the swing. We want to use energy conservation. The tension T is perpendicular to the direction of motion, thus does no work; air resistance is negligible; only gravity does work. E 1 = E 2, K 1 + U 1 = K 2 + U 2 (20) Since velocity is 0 at the top of the swing, and gravitational potential energy is so U 1 = mgl(1 cos θ), (21) mgl(1 cos θ) = 1 2 mv2 2 (22) v 2 = 2gl(1 cos θ) (23)

5 8.01L Problem Set #8 Solutions, p. 5 (b) What is the tension in the string when it makes an angle of θ with the vertical (i.e., when it is at its maximum angle)? At θ from the vertical, the rock is at the top of the swing, thus has 0 speed. Apply F = m a to the radial direction T mg cos θ = 0. The tension in the spring is T = mg cos θ (24) (c) What is the tension in the string as it passes through the vertical? We now consider when it passes through the vertical (point 2). Again, let s apply Newton s 2nd law. Fy = ma y (25) Thus The tension increases from point 1 to 2. T mg = m v2 2 l. (26) T = mg + m v2 2 l (27) T = mg(3 2 cos θ) (28) 8-4: Potential Energy Young and Freedman (13th edition) problem 7.86, page 239. (same as Young and Freedman (12th edition) problem 7.86, page 246.) A particle moves along the x-axis while acted on by a single conservative force parallel to the x-axis. The force corresponds to the potential-energy function graphed in Fig. P7.86. The particle is released from rest at point A. (a) What is the direction of the force on the particle when it is at point A? F (x A ) = du x=xa = positive, i.e. rightward. (b) At point B? dx Figure P7.86

6 8.01L Problem Set #8 Solutions, p. 6 F (x B ) = du dx x=xb = negative, i.e. leftward. (c) At what value of x is the kinetic energy of the particle a maximum? E = K + U is fixed, so K is maximum when U is minimized x 0.75 m. (d) What is the force on the particle when it is at point C? F (x C ) = du x=xc = 0. dx (e) What is the largest value of x reached by the particle during its motion? x max is the value of x when all energy is potential again, i.e. when a horizontal line through the starting point A intersects blue curve. That is x max 2.1 m. (f) What value or values of x correspond to points of stable equilibrium? The minima of U are stable equilibrium: (g) Of unstable equilibrium? x 0.75 m, and (29) x 1.8 m. (30) Maxima & saddle points of U are unstable equilibria x 1.4 m (31) 8-5: Force of a Baseball Swing Adapted from Young and Freedman (13th edition) exercise 8.8, page 269. (a) A baseball has mass m b. If the velocity of a pitched ball has a magnitude of v i and the batted ball s velocity is v f in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. The initial and final momentum (x-component) of the ball are p 1,x = m b v i (32) p 2,x = m b v f (33)

7 8.01L Problem Set #8 Solutions, p. 7 Apply impulse-momentum theorem, J = p 2 p 1, the x-component of impulse equals the change in the x-momentum J x = p 2x p 1x = m(v f + v i ) (34) Therefore, the magnitude of the change in momentum of the ball is J x = m(v f + v i ) which is equal to the magnitude of the impulse applied to it by the bat. (b) If the ball has a mass of kg, an initial speed of 45.0 m/s, a final speed of 55.0 m/s, and the ball remains in contact with the bat for 2.00 ms (milliseconds), find the magnitude of the average force applied by the bat. (F av ) x = J x t The collision time is t = 2.00 ms = s. From J x = (F av ) x t, = (0.145 kg)(45.0 m/s m/s) s = 14.5 N s s = 7250 N (35) Thus the bat applies an average force of 7250 N to the ball in the x-direction. 8-6: Throwing a Rock Adapted from Young and Freedman (13th edition) exercise 8.28, page 270. (a) You are standing on a large sheet of frictionless ice and holding a large rock. In order to get off the ice, you throw the rock so it has velocity v r relative to the earth at an angle θ above the horizontal. If your mass is M and the rock s mass is m r, what is your speed after you throw the rock? Consider the rock and yourself together as one system. Then since there is no horizontal external force acting on this system, the x-component of the momentum of the system is conserved, p x1 = p 2x. Since initially you and the rock are at rest, p 1x = 0. After you throw away the rock, the total momentum (x-component) is the sum of the momentum of you and the rock.

8 8.01L Problem Set #8 Solutions, p. 8 p 2x = m r v rx + m p v px where m r, m p are the masses of the rock and you, respectively, and v rx, v px are the x-components of the velocities. v px = p 2x m r v rx m p = m r m p v rx = m r m p v r cos θ. (36) (b) Calculate for the following values: v r = 12.0 m/s, θ = 35.0, M = 70.0 kg, m r = 15.0 kg. v rx = v r cos 35 = 12.0 m/s cos 35 = 9.83 m/s (37) v px = 15.0 kg 9.83 m/s = 2.11 m/s 70.0 kg (38) (a negative sign means you are moving in the opposite direction of the rock). Since you are confined to move on the horizontal surface v p = v px î = (2.11 m/s) î (39) Your speed after throwing the rock is 2.11 m/s. 8-7: Speed of a Bullet Adapted from Young and Freedman (13th edition) exercise 8.42, page 271. (a) A bullet with mass m b is fired horizontally into a wooden target block of mass M t resting on a horizontal surface. The coefficient of kinetic friction between block and surface is µ. The bullet remains embedded in the block, which is observed to slide a distance d along the surface before stopping. What was the initial speed of the bullet? Set the initial speed of the bullet to be v b. Consider the bullet and the block together as the system. Assume during the first stage, the bullet embeds itself in the block so quickly that the impulse due to the friction between the surface and block (external force) is negligible. Then we can apply momentum conservation to find the velocity of the system after the bullet embeds in the block and they

9 8.01L Problem Set #8 Solutions, p. 9 move together with the same speed v 1. By momentum conservation, p 1 = p 0. p 1 = (m b + m)v 1, p 0 = m b v b (40) v 1 = m bv b (41) m b + m For the second stage, we use the work-energy theorem. (since the potential energy doesn t change). Final speed is 0 K = 1 2 mv2 K = W f (42) W f = fd, f = µn = µmg (43) 1 2 mv2 1 = µmgd (44) ( mb v b m b + m v1 2 ) 2 = 2µdg (45) = 2µgd (46) v b = 2µgd m b + m m b (47) (b) Calculate using m b = 5 g, M t = 1.20 kg, µ = 0.20, d = m. v b = 2µgd m b + m (48) m b = m/s m kg kg (49) kg = 229 m/s (50) Therefore the initial speed of the bullet is 229 m/s. 8-8: The Roller Skater and the Football BCG problem 5B.4, page 175. This problem has a hint on page 194, an answer to the hint on page 196, and an answer to the problem on page 197. You are roller-skating peacefully down a street at a constant speed of 3 m/s when someone suddenly throws a football at you from directly ahead. If your mass is 65 kg and the football s is 0.40 kg, what is your speed afterwards if (a) you catch the ball, which was thrown at you with a horizontal velocity of 15 m/s; (b) you miss it and it bounces off you with a velocity of 10 m/s (relative to the street) in the opposite direction? In each case, what is the total kinetic energy of the system comprising you and the ball before and after your interaction? (Remember: solve the answer in terms of variables before you start sticking in numbers!)

10 8.01L Problem Set #8 Solutions, p. 10 In this problem we will use conservation of momentum! Let s name our variables first. Let the mass of the person be M, the mass of the football be m. The initial velocity of the person is v and that of the ball v b. The final speed of the person is v f and if the ball is not caught by the person, the speed will be v b,2 but of course if the person catches the ball, the latter will have the same speed v f as the skater. A good habit is to write down the momenta before and after the collision. For the first case, quantity Before collision After collision p x Mv mv b (M + m)v f 1 KE 2 Mv mv2 b = J 1 2 m)v2 f By conservation of momentum, in the first case, the final velocity is: v f = Mv mv b M + m = 65 kg 3 m/s 0.4 kg 15 m/s 65 kg kg = 2.89 m/s (51) For the second case: quantity Before collision After collision p x Mv mv b Mv f + mv 2 1 KE 2 Mv mv2 b = J 1 2 Mv2 f mv2 2 v f = Mv m(v b + v 2 ) M = 65 kg 3 m/s 0.4 kg(15 m/s + 10 m/s) 65 kg = 2.85 m/s (52) Knowing these values for v f, we can calculate the final kinetic energy for each of these cases: Case 1: K f = 1 2 (M + m)v2 f = J (53) Case 2: K f = 1 2 Mv2 f mv2 2 = J. (54)

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