Question

A canon ball is fired with a velocity $200m/sec$ at an angle of ${60}^{°}$ with the horizontal. At the highest point of its flight it explodes into $3$ equal fragments, one going vertically upwards with a velocity $100m/sec$, the second one falling vertically downwards with a velocity $100m/sec$. The third fragment will be moving with a velocity,

• A. $100m/s$ in the horizontal direction
• B. $300m/s$ in the horizontal direction
• C. $300m/s$ in a direction making an angle of $60°$ with the horizontal
• D. $200m/s$ in a direction making an angle of ${60}^{°}$ with the horizontal

Answer 1

The correct option is C

$300m/s$ in a direction making an angle of $60°$ with the horizontal

Step 1: Given data and drawing the diagram

The velocity of ball, $u=200m/s$

Angle with horizon $={60}^{°}$

The momentum of fragments $=\overrightarrow{P_1},\overrightarrow{P_2},\overrightarrow{P_3}$

Let the velocity of a third part after collision $=V$

Step 2: Find the momentum

The momentum of the ball (mass $m$) before explosion at the highest point,

$$\begin{gathered} =mv\hat{i} \\ =mu(\cos {60}^{°})\hat{i} \\ =m\times 200\times 0.5\hat{i} \\ =100m\hat{i} \end{gathered}$$

[Since, at the maximum height, the initial velocity have only horizontal component as the vertical component becomes zero]

After the collision, momentum is,

$=\overrightarrow{P_1}+\overrightarrow{P_2}+\overrightarrow{P_3}$

$=\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}\times V\hat{i}$

[According to the law of conservation of linear momentum]

Step 3: Find the velocity of third fragment

According to law of momentum,

Momentum before collision = momentum after collision

$100m\hat{i}=\frac{m}{3}\times 100\hat{j}-\frac{m}{3}\times 100\hat{j}+\frac{m}{3}\times V\hat{i}$

$\Rightarrow$ $V=300m/s$

Hence, option (C) is correct.