Ejercicios De Fisica Fundamental

Chapter 9. Impulse and Momentum

9-1. A 0.5-kg wrench is dropped from a height of 10 m. What is its momentum just before it strikes the floor? (First find the velocity from conservation of energy.)
mgh = ½mv2; [pic] v = 14.0 m/s
p = mv = (0.5 kg)(14 m/s); p = 7.00 kg m/s, down

9-2. Compute the momentum and kinetic energy of a 2400-lb car moving north at 55 mi/h.
[pic]v = 55 mi/h = 80.7 ft/s
p = mv = (75 slugs)(80.7 ft/s); p = 6050 slug ft/s
K = ½mv2 = ½(75 slugs)(80.66 ft/s)2; K = 244,000 ft lb

9-3. A 2500-kg truck traveling at 40 km/h strikes a brick wall and comes to a stop in 0.2 s. (a) What is the change in momentum? (b) What is the impulse? (c) What is the average force on the wall during the crash? Take + to be toward the wall.( 40 km/h = 11.1 m/s)
(p = mvf – mvo = 0 - (2500 kg)(11.1 m/s); (p = - 27,800 kg m/s
Impulse = (p; F t = -27,800 kg m/s
Force ON truck: [pic] F = -139,000 N
Force on wall is opposite, so F = + 139,000 N

9-4. What is the momentum of a 3-kg bullet moving at 600 m/s in a direction 300 above the horizontal? What are the horizontal and vertical components of thismomentum?
p = mv = (3 kg)(600 m/s); p = 1800 kg m/s, 300
px = 1800 cos 300 and py = 1800 sin 300; px = 1560 kg m/s; py = 900 kg m/s
*9-5. A 0.2-kg baseball traveling to the left at 20 m/s is driven in the opposite direction at 35 m/s when it is hit by a bat. The average force on the ball is 6400 N. How long was it in contact with the bat? (Impulse = change in momentum. )
F Δt = mvf – mvo = (0.2 kg)(35 m/s) – (0.2 kg)(-20 m/s)
(6400 N) Δt = 11 kg m/s; Δt = 1.72 ms

*9-6. A bat exerts an average force of 248 lb on 0.6-lb ball for 0.01 s. The incoming velocity of the ball was 44 ft/s. If it leaves in the opposite direction what is its velocity?
Choose positive + as direction away from the bat,making incoming ball velocity negative:
F Δt = mvf – mvo ; F Δt = mvf – mvo ; [pic]
(240 lb)(0.01 s) = (0.01875 slugs)vf - (0.01875 slugs)(-44 ft/s)
0.01875 vf = 2.4 lb s – 0.825; vf = 84.0 ft/s

*9-7. A 500-g ball travels from left to right at 20 m/s. A bat drives the ball in the opposite direction with avelocity of 36 m/s. The time of contact was 0.003 s. What was the average force on the ball? ( m = 0.5 kg, vo = +20 m/s, vf = -36 m/s, Δt = 0.003 s )
F Δt = mvf – mvo ; F(0.003 s) = (0.5 kg)(-36 m/s) – (0.5 kg)(20 m/s)
[pic]; F = 9330 N
*9-8. A 400-g rubber ball is dropped a vertical distance of 12 m onto the pavement. It is in contact with the pavement for 0.01 s and reboundsto a height of 10 m. What is the total change in momentum? What average force is exerted on the ball?
To apply the impulse-momentum theorem, we need to first find the velocities just before and just after impact with the ground.
(Ep)Beginning = (Ek)Ground ; mgho = ½mvo2;
[pic] vo = - 15.3 m/s
½mvf2 = mghf ; [pic] vf = + 14 m/sFΔt = mvf – mvo; F(0.01 s) = (0.4 kg)(14 m/s) – (0.4 kg)(-15.3 m/s); F = 1170 N

*9-9. A cue stick strikes an eight-ball with an average force of 80 N over a time of 12 ms. If the mass of the ball is 200 g, what will be its velocity?
FΔt = mvf – mvo; (80 N)(0.012 s) = (0.2 kg)vf – 0; v = 4.80 m/s

9-10. A golfer hits a 46 g golf ball with an initial velocity of 50m/s at 300. What are the x- and y-components of the momentum imparted to the ball?
vx = (50) cos 300 = 43.3 m/s ; vy = (50) sin 300 = 25.0 m/s
px = (0.046 kg)(43.3 m/s); py = (0.046 kg)(25 m/s) px = 1.99 kg m/s; py = 1.15 m/s

*9-11. The face of the club in Problem 9-10 is in contact with the ball for 1.5 ms. What are the horizontal and vertical components of the...