Question

A 7.0−kg object is at rest on a perfectly frictionless surface when it is struck head-on by a 3.5−kg object moving at 13 m/s. If the collision is perfectly elastic, what is the speed of the 7.0−kg object after the collision? [Hint: You will need two equations.]

Answer 1

4.72 m/s

Step-by-step explanation:

Since the collision is perfectly elastic, both momentum and energy is conserved

Momenum:
`mv_1 = mv_2 + MV`

Kinetic energy:
`0.5mv_1^2 = 0.5mv_2^2 + 0.5MV^2`

where m = 3.5 kg is the mass of the small moving object at speed v1 = 13 m/s. M = 7 kg is the mass of the big collided object with speed V post-collision. v2 is the speed of the small object post-collision.

Momentum:
`3.5*13 = 3.5v_2 + 7V`

`13 = v_2 + 2V`

`v_2 = 13 - 2V`

Energy:
`3.5*13^2 = 3.5v_2^2 + 7V^2`

`169 = v_2^2 + 2V^2`

We can substitute
`v_2 = 13 - 2V`

`169 = (13 - 2V)^2 + 7V^2`

`169 = 169 - 52V + 4V^2 + 7V^2`

`V(11V - 52) = 0`

`11V - 52 = 0`

`V = 52/11 = 4.72 m/s`