Question

A cannon ball of mass 2 kg is fired at a target 500 meters away. The 400 kg cannon recoils with a velocity of $2m{s}^{-1}$. Assuming cannon ball moves with constant velocity, when will it hit the target?

• A. 1.25 sec
• B. 2 sec
• C. 1.5 sec
• D. 0.2 sec

Answer 1

The correct option is A

1.25 sec

Here, mass of cannon ball $m_1$= 2 kg,

velocity of cannon after firing $v_2$= $2m{s}^{-1}$,

mass of cannon $m_2$ = 400 kg,

velocity of cannon before firing $u_2$ = $0m{s}^{-1}$,

velocity of cannon ball before firing $u_1$ = $0m{s}^{-1}$ and velocity of cannon ball after firing = $v_1$

Total momentum before firing = $m_1\times u_1+m_2\times u_2=0kgm{s}^{-1}$ , Total momentum after firing = $m_1\times v_1+m_2\times v_2$

From the law of conservation of momentum, Total momentum before firing = Total momentum after firing. Therefore,
0 = $m_1\times v_1+m_2\times v_2$
$0=400\times 2+2\times v_2$
Hence, the velocity of the cannonball $v_2$ will be 400 $m{s}^{-1}$.
Thus, time taken to hit the target= $\frac{500}{400}$ = 1.25 s