Question

A 4.0kg bowling ball sliding to the right at 8.0 m/s has an elastic head-on collision with another 4.0 kg bowling ball initially at rest. The first ball stops after the collision

Answer 1

a)

We use the formula :

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values in:

4.0kg*8.0m/s + 4.0kg*0m/s = 4.0kg*0m/s +4.0kg*v2f

Calculating this we get:

32.0kg*m/s + 0kg*m/s = 0kg*m/s + 4.0kg*v2f

Rearrange for v2f:

v2f = $\frac{32.0kg*m/s}{4.0kg}$

This gives us 8.0 m/s as the final velocity of the second ball.

b)

Since the collision is assumed to be elastic it means that the kinetic energy must be equal before and after the collision.

This means we use the formula:

Ek = $\frac{1}{2} *m*v^{2}$+ $\frac{1}{2} *m*v^{2}$ = $\frac{1}{2} *m*v^{2}$ +  $\frac{1}{2}*m*v^{2}$

Substituting in values:

Ek = 0.5*4.0kg*(8.0m/s)^2 + 0.5*4.0kg*(0m/s)^2 = 0.5*4.0kg*(0m/s)^2 + 0.5*4.0kg*(8.0m/s)^2

This simplifies to:

Ek= 128J + 0J = 0J + 128J

This shows us that the kinetic energy is equal on each side therefore the collision is Elastic and no energy has been lost.