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 will it hit the target?

• A. $1.25s$
• B. $2s$
• C. $1.5s$
• D. $0.2s$

Answer 1

The correct option is A

$1.25s$

Let mass of cannon be $m_1$ = $400kg$
Let mass of cannon ball be $m_2$ = $2kg$
Let initial velocities of the cannon and cannon ball be $u_1$ and $u_2$ respectively.
Let final velocity of cannon and cannon ball be $v_1$ and $v_2$ respectively.
From the law of conservation of momentum,
total momentum before firing = total momentum after firing.
$\therefore$ $m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$
As the cannon and cannon ball is at rest initially, the initial velocity will be zero, i.e
$0=m_1{v}_1+m_2{v}_2$
Substituting the values:
$0=400\times (-2)+2\times v$

$\Rightarrow v=\frac{400\times 2}{2}=400m{s}^{-1}$
Hence, the velocity of the cannonball will be $400m{s}^{-1}$​
Since we know,
$time=\frac{displacement}{velocity}$
Thus, time taken to hit the target= $\frac{500}{400}$ = $1.25s$