Question

A shell is fired from a cannon with a speed of 100 m/s at an angle ${60}^{\circ }$ with the horizontal (positive x - direction). At the highest point of its trajectory, the shell explodes into two equal fragments. One of the fragments moves along the negative x-direction with a speed of 50 m/s. What is the speed of the other fragment at the time of explosion.

• A. 100 m/s
• B. 200 m/s
• C. 150 m/s
• D. 300 m/s

Answer 1

The correct option is C 150 m/s

Momentum conservation technique is used to solve these kinds of problems.

Let's assume that the horizontal component of the velocity the other fragment is Vx and the vertical component is Vy.

Initial momentum in X direction = Final momentum in X direction

$\left(m\right)\left(100cos{60}^0\right)\left(\hat{i}\right)$= ($\frac{m}{2}$)($50$) $(-\hat{i})$ + ($\frac{m}{2}$)($V_H$) $\left(\hat{i}\right)$

$V_H$ = $150$ $\left(\hat{i}\right)$ m/s

Similarly in Y direction

Initial momentum in Y direction = Final momentum in Y direction

$\left(m\right)\left(100sin{60}^0\right)\left(\hat{j}\right)$ = ($\frac{m}{2}$)($V_V$) $\left(\hat{j}\right)$

$V_V$ = $100\sqrt{(}3)$ $\left(\hat{j}\right)$ m/s

Velocity of the other fragment = $\sqrt{(}{V}_V^2+V_H^2)$

= $\sqrt{(}100\sqrt{3}{)}^2+{150}^2)$

= 229 m/s