Question

WILL MARK AS BRAINLIEST!!!!!!


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

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