Question

A 0. 0427 kg racquetball is moving 22. 3 m/s when it strikes a stationary box. The ball bounces back at 11. 5 m/s, while the box moves forward at 1. 53 m/s. What is the mass of the box?.

Answer 1

To solve the question we must know about the conservation of momentum.

Conservation of Momentum

According to the law of conservation of momentum, when two bodies collide with each other then their momentum is conserved if no external force acts on the system. therefore, the combined momentum of the two objects before collision will be equal to the combined momentum of the two objects after the collision.

the momentum of the two objects before the collision

                       = momentum of the two objects after the collision

The mass of the box is 0.30141 kg.

Explanation

Given to us

• Mass of racquetball, m = 0.0427 kg,
• Velocity of racquetball before Collision,  $\bold{u_1}$ = 22.3 m/s,
• Velocity of box before Collision,$\bold{ u_2}$  = 0 m/s,
• Velocity of racquetball after Collision, $\bold{v_1}$ = 11.5 m/s,
• Velocity of box after Collision, $\bold{v_2}$ = 1.53 m/s,

Assumption

Let the mass of the box be M.

Conservation Of Momentum

Momentum before collision = Momentum after collision

$mu_1 + Mu_2 = mv_1 + Mv_2$

given, the box was stationary before the collision,

$mu_1 + Mu_2 = mv_1 + Mv_2\\\\ mu_1 + 0 = mv_1 + Mv_2\\\\ mu_1 - mv_1 = Mv_2\\\\ M= \dfrac{mu_1 - mv_1}{v_2}$

Substituting the values,

$M= \dfrac{mu_1 - mv_1}{v_2}\\\\ M = \dfrac{(0.0427 \times 22.3) - (0.0427\times 11.5)}{1.53}\\\\ M = 0.30141\ kg$

Hence, the mass of the box is 0.30141 kg.

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