1. A 0.2 kg rubber ball is dropped from the window of
a building. It strikes the sidewalk below at 30 m/s at an angle of
30o and rebounds up at 20 m/s at the same angle. The magnitude
of the impulse (in N.s) due to the collision with the sidewalk is:
A) (+10, - 8.6) B) (+10, +8.6)
C) ( - 1.7, +5)
D) ( - 1.7, - 0.2)
E) (+5.0, -0.2)
Impulse = momentum change = final - initial = 0.2*20(cos30,sin30) - 0.2*30(cos30,-sin30) = (-1.7,5)
2. A 5-ton freight car, moving at 4 mph, collides and
couples with an 8-ton freight car, which was initially at rest. The
common final speed of these two cars is:
A) 1 mph B) 1.31
mph C) 1.54 mph
D) 2.5 mph E) 4 mph
conservation of momentum initial = 5*4 + 0 = final = (5+8)*v so v = 20/13 = 1.54
3. A 5-g bullet is fired horizontally into a 10-kg block of wood suspended by a rope from the ceiling. The block swings in an arc, rising 5 mm above its lowest position. The velocity of the bullet was:
A) Need to know how much heat was generated in the collision
B) 2400 cm/s
C) 2.4*106 cm/s
D) 800 cm/s
E) 8*104 cm/s
a) conservation of momentum initial
= 0.005*v + 0 = final = (0.005 + 10)*vB
b) conservation of energy for block moving up by h = 5 mm:
1/2 (0.005 + 10) vB2 = (0.005 + 10)gh = (0.005 +
10)*9.8*0.005
then from b) vB = sqrt(2*9.8*0.005) =
sqrt(0.098) = 0.313
then from a) 0.005*v = 10.005*0.313 and
v = 626 m/s The closest answer is E