Question

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

• A. 5 sec
• B. 10 sec
• C. 15 sec
• D. 2 sec

Answer 1

The correct option is A

5 sec

Given: mass of cannon, $m_1=400kg$, mass of cannonball, $m_2=2kg$, cannon recoil velocity, $v_1=2m{s}^{-1}$ (The negative sign indicates that the recoil velocity is opposite to the direction of cannon velocity.), target distance $s=500m$
Let the speed of the cannonball be $v_2$. Let the time taken to hit the target be $t$.

Total initial momentum $=0$ (as both the cannon and the cannonball are at rest)
Total final momentum $=m_1{v}_1+m_2{v}_2$

From the law of conservation of momentum,
Total momentum before firing = Total momentum after firing
$\Rightarrow 0=m_1{v}_1+m_2{v}_2$

$\Rightarrow 0=(400\times (-2\left)\right)+(2\times v_2)$

$\Rightarrow v_2=\frac{400\times 2}{2}=400m{s}^{-1}$
Hence, the velocity of the cannonball is $400m{s}^{-1}$.

Time $t=\frac{s}{v_2}=\frac{500}{400}=1.25s$
Thus, the time taken to hit the target$=1.25s$.