Question

A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the heavier ball moves at 2m/s in the opposite direction

Answer 1

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

$m_{1}$$u_{1}$ + $m_{2}$ $u_{2}$ = $m_{1}$$v_{1}$ - $m_{2}$$v_{2}$

$m_{1}$ = 3 kg, $u_{1}$ = 8 m/s, $m_{2}$ = 2 kg, $u_{2}$ = 0 m/s ( since it is at rest), $v_{1}$ = 2 m/s, $v_{2}$ = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x $v_{2}$)

24 + 0 = 16 - 2$v_{2}$

2$v_{2}$ = 16 - 24

2$v_{2}$ = -8

$v_{2}$ = $\frac{-8}{2}$

   = -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.