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

$1.25 s$

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$2 s$

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$1.5 s$

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$0.2 s$

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Solution

The correct option is A
$1.25 s$

Let mass of cannon be $m_{1}$ = $400 k g$
Let mass of cannon ball be $m_{2}$ = $2 k g$
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.
$∴$ $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
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:
$0 = 400 \times (- 2) + 2 \times v$

$\Rightarrow v = \frac{400 \times 2}{2} = 400 m s^{- 1}$
Hence, the velocity of the cannonball will be $400 m s^{- 1}$
Since we know,
$t i m e = \frac{d i s p l a c e m e n t}{v e l o c i t y}$
Thus, time taken to hit the target= $\frac{500}{400}$ = $1.25 s$

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