Question

A steel ball of mass 0.500 kgis fastened to a cord that is 70.0cmlong and fixed at the far end. The ball is then released when the cord is horizontal (Fig. 9-65). At the bottom of its path, the ball strikes a 2.50 kgsteel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

Short answer

1. The speeds of the ball is,$V_{1f}=-2.47\mathrm{m}/\mathrm{s}$
2. The speed of the block is,$V_{2f}=1.23\mathrm{m}/\mathrm{s}$

Step-by-step solution

Step 1: Given Data

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, $m_2=2.50\mathrm{kg}$

Determining the concept

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.

(a) Determining the speed of the ball just after the collision

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

The mechanical energy conversion,

Potential energy = kinetic energy

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

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

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

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

(b) Determining the speed of the block just after the collision

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

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

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.