Question

Mike shoots a large marble (Marble A, mass: 0.05 kg) at a smaller marble (Marble B, mass: 0.03 kg) that is sitting still. Marble A was initially moving at a velocity of 0.6 m/s, but after the collision it has a velocity of –0.2 m/s. What is the resulting velocity of marble B after the collision? Be sure to show your work for solving this problem along with the final answer.

Answer 1

Since no angle is given, then we will assume linear collision.

Now, based on the law on conservation of momentum, the total momentum before collision will be equal to the total momentum after collision.

This means that:
M1V1 + M2V2 = M1V1' + M2V2'
where:
M1 is the mass of the large marble = 0.05 kg
V1 is the initial velocity of the large marble = 0.6 m/sec
M2 is the mass of the small marble = 0.03 kg
V2 is the initial velocity of the small marble = 0 m/sec (marble is at rest)
V1' is the final velocity of the large marble = -0.2 m/sec
V2' is the final velocity of the small marble that we want to calculate

Substitute with the givens in the above equation to get V2' as follows:
M1V1 + M2V2 = M1V1' + M2V2'
(0.05)(0.6) + (0.03)(0) = (0.05)(-0.2) + 0.03V2'
0.03 = -0.01 + 0.03V2'
0.03V2' = 0.03+0.01 = 0.04
V2' = 0.04/0.03
V2' = 1.334 m/sec