At the same instant that a 0.50-kg ball is dropped from 25 m above Earth, a second ball, with a mass of 0.25 kg, is thrown straight upward from Earth's surface with an initial speed of 15 m/s.They move along nearby lines and pass without colliding.At the end of 2.0 s the velocity of the center of mass of the two-ball system is:
A)11 m/s, down
B)11 m/s, up
C)15 m/s, down
D)15 m/s, up
E)20 m/s, down

Question

Quizplus · Accepted Answer

First, we can find the initial momentum of the system, which is zero since the balls are initially at rest (the upward motion of the second ball doesn't affect the momentum because we are considering the center-of-mass motion of the system, which depends only on the total mass of the system and not on the internal motions of its parts).

At the end of 2.0 s, the first ball will have fallen a distance of:

y = (1/2)gt^2 = (1/2)(9.81 m/s^2)(2.0 s)^2 = 19.62 m

where g is the acceleration due to gravity. The speed of the first ball just before it hits the ground is:

v1 = √(2gy) = √(2(9.81 m/s^2)(19.62 m)) ≈ 19.8 m/s

The second ball will rise to a maximum height of:

ymax = (1/2)gt^2 = (1/2)(9.81 m/s^2)(2.0 s)^2 = 19.62 m

and then fall back down to Earth with the same speed that it had initially, 15 m/s. So at the end of 2.0 s, the velocity of the second ball is:

v2 = -15 m/s

where the negative sign indicates that it is moving downward.

The final momentum of the system is the sum of the momenta of the two balls:

p = m1v1 + m2v2 = (0.50 kg)(19.8 m/s) + (0.25 kg)(-15 m/s) ≈ 6.85 kg m/s

where the positive direction is downward. The final velocity of the center of mass of the system is:

vcm = p / (m1 + m2) = 6.85 kg m/s / (0.50 kg + 0.25 kg) ≈ 15 m/s

where the positive direction is again downward. Therefore, the answer is (C) 15 m/s, down.

At the Same Instant That a 0