Question

A 44.0 g marble moving at 1.90 m/s strikes a 22.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.

What is the speed of 44.0 g marble immediately after the collision?

What is the speed of 22.0 g marble immediately after the collision?

Answer 1

Answer: Here it is showing conservation of momentum and energy.

Given that 44 g marble and 22 g marble and velocity=1.9 m/s

We know that for an elastic, head-on collisions the relative velocity of approach = relative velocity of separation,

Given relative velocity of approach, 44 g marble=1.9 m/s.

Therefore

1.90 m/s = v - u
where u, v are the after-collision velocities of 44 g marble and 20 g marble, respectively
So v = u + 1.90m/s

Now we have to apply Conservation of momentum

44 g x 1.90m/s + 0 = 44 g x u + 22 g x v

But v=u+1.90

83.6 = 44u + 22(u + 1.90) = 66u + 41.8

u = 41.8/ 66 = 0.633 m/s for 44.0 g marble
v = u + 1.90m/s =0.633 + 1.9 = 2 53 m/s for 20.0 g marble.

Please let me know if you have any doubt. Thanks.