Question

A 0.0427 kg racquet-ball is moving

Answer 1

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg

Velocity of racket-ball before collision $(v_{1i})$ = 22.3 m/s

Velocity of racket-ball after collision with box $(v_{1f})$ = -11.5 m/s

[Since ball is bouncing back, so velocity is taken negative.]

Velocity of the box before collision $v_{2i}$ = 0 m/s

[Since the box is stationary, so velocity is taken zero]

Velocity of box moving forward after collision $v_{2f}$= 1.53 m/s

To find the mas of the box $m_2$.

By law of conservation of momentum we have:

Momentum before collision = Momentum after collision

This can be written as:

$p_i=p_f$

$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$

We can plugin the given value to find $m_2$

$(0.0427\times 22.3)+(m_2\times 0)=(0.0427\times (-11.5))(m_2\times 1.53)$

$0.9522+0=-0.4911+1.53m_2$

Adding both sides by 0.4911

$0.9522+0.4911=-0.4911+0.4911+1.53m_2$

$1.4433=1.53m_2$

Dividing both sides by 1.53.

$\frac{1.4433}{1.53}=\frac{1.53m_2}{1.53}$

$0.9433=m_2$

∴ $m_2=0.9433$ kg

Mass of the box = 0.9433 kg (Answer)