Problem#1
High-speed stroboscopic photographs show that the head of a golf club of mass 200 g is traveling at 55.0 m/s just before it strikes a 46.0-g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40.0 m/s. Find the speed of the golf ball just after impact.
Answer:
the speed of the golf ball just after impact is
m1v1i + m2v2i = m1v1f + m2v2f
(200 g)(55.0 m/s) + 0 = (200 g)(40.0 m/s) + (46.0 g)v
v = 65.2 m/s
Problem#2
An archer shoots an arrow toward a target that is sliding toward her with a speed of 2.50 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 35.0 m/s and passes through the 300-g target, which is stopped by the impact. What is the speed of the arrow after passing through the target?
Answer:
We use
m1v1i + m2v2i = m1v1f + m2v2f
(22.5 g)(35.0 m/s) + (300 g)(–2.50 m/s) = (22.5 g)vif + 0
v1f = (37.5 g m/s)/(22.5 g) = 1.67m/s
Problem#3
A 10.0-g bullet is fired into a stationary block of wood (m = 5.00 kg). The relative motion of the bullet stops inside the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.600 m/s. What was the original speed of the bullet?
Answer:
Momentum is conserved
mbv = mwvw
(10.0 x 10-3 kg)v = (5.00 kg)(0.600 m/s)
v = 301 m/s
Problem#4
A railroad car of mass 2.50 x 104 kg is moving with a speed of 4.00 m/s. It collides and couples with three other coupled railroad cars, each of the same mass as the single car and moving in the same direction with an initial speed of 2.00 m/s. (a) What is the speed of the four cars after the collision? (b) How much mechanical energy is lost in the collision?
Answer:
(a) the speed of the four cars after the collision is
mv1i + 3mv2i = 4mv
4.00 m/s + 3(2.00 m/s) = 4v
v = 2.50 m/s
(c) mechanical energy is lost in the collision is
∆K = Kf – Ki
∆K = ½ (2.50 x 104 kg)[(4.00 m/s)2 – (2.00 m/s)2] = –37.5 kJ
Problem#5
Four railroad cars, each of mass 2.50 x 104 kg, are coupled together and coasting along horizontal tracks at speed vi toward the south. A very strong but foolish movie actor, riding on the second car, uncouples the front car and gives it a big push, increasing its speed to 4.00 m/s southward. The remaining three cars continue moving south, now at 2.00 m/s. (a) Find the initial speed of the
cars. (b) How much work did the actor do? (c) State the relationship between the process described here and the process in Problem 4.
Answer:
(a) The internal forces exerted by the actor do not change the total momentum of the system of the four cars and the movie actor
4mvi = 3m(2.00 m/s) + m(4.00 m/s)
4vi = 10.0 m/s
vi = 2.50 m/s
(b) work did the actor do is
W = ∆K
W = ½ [3m(2.00 m/s)2 + m(4.00 m/s)2] – ½ (4m)(2.50 m/s)2
W = ½ (2.50 x 104 kg)(12.0 + 16.0 – 25.0) m2/s2 = 37.5 kJ
(c) The event considered here is the time reversal of the perfectly inelastic collision in the previous
problem. The same momentum conservation equation describes both processes.
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