91Ó°ÊÓ

Mass of ball,$m_1=0.500\mathrm{kg}$

Length of cord having ball attached is,$h=70\mathrm{cm}=0.7\mathrm{m}$

Mass of block is, $m_2=2.50\mathrm{kg}$

By using the concept of the law of conservation of linear momentum, it can be said that momentum before collision is equal to the momentum after collision. Also, the law of conservation of mechanical energy can be used to find the speed of the object after the collision.

Formulae are as follow:

For conservation of linear momentum $m_1{V}_{1i}+m_2{V}_{2i}=m_1{V}_{1f}+m_2{V}_{2f}$

Speed in center of mass is,role="math" localid="1661575299217" $V_{com}=\frac{m_1{V}_{1i}+m_2{V}_{2i}}{m_1+m_2}$

For elastic collision,role="math" localid="1661575306077" $V_{1f}=\frac{m_1-m_2}{m_1+m_2}V$

$V_{2f}=\frac{\left(2m_1\right)}{m_1+m_2}V$

Where, m₁, m₂ are masses and V is velocity.

First, find the speed Vof the ball of mass$m_1$right before the collision.

The mechanical energy conversion,

Potential energy = kinetic energy

$$\begin{gathered} m_{1}gh=\frac{1}{2}{m}_{!}{V}^2 \\ V=\sqrt{2gh} \\ V=\sqrt{2×9.50×0.700} \\ V=\sqrt{13.3} \\ V=3.7\mathrm{m}/\mathrm{s} \end{gathered}$$

Now, treat the elastic collision using equations (9 – 67),

$$\begin{gathered} V_{1f}=\frac{m_1-m_2}{m_1+m_2}V \\ {V}_{1f}=\frac{0.5-2.5}{0.5+2.5}×3.7 \\ {V}_{1f}=\frac{-2}{3}×3.7 \\ {V}_{1f}=-2.47\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speeds of the ball is,$V_{1f}=-2.47\mathrm{m}/\mathrm{s}$

Use equations (9 – 68) to find the final speed of the block,

$$\begin{gathered} V_{2f}=\frac{\left(2m_1\right)}{m_1+m_2}V \\ {V}_{2f}=\frac{2×0.5}{0.5+2.5}×3.7 \\ {V}_{2f}=\frac{1}{3}×3.7 \\ {V}_{2f}=1.23\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence,the speed of the block is,$V_{2f}=1.23\mathrm{m}/\mathrm{s}$

Therefore, by using the concept of conservation of mechanical energy, the speed of the ball and block can be found.