Question

# A cannon ball (mass = $2kg$) is fired at a target of $500m$ away. The $400kg$ cannon recoils by $50cm$. Assuming the cannon ball moves with constant velocity, when will it hit the target? [Resistance offered by ground to cannon is $100N$]

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Solution

## Step 1. Given data:Mass of cannon ball $\left(m\right)$ = $2kg$Mass of cannon $\left(M\right)$ = $400kg$Distance of target $\left(D\right)$ = $500m$Recoiling of cannon $\left(s\right)$= $50cm$Resistance offered by ground to cannon $\left(F\right)$ = $100N$Step 2. Formula used:Since no external force is applied to the system, therefore, the momentum of the system is conserved and hence we get $mv=-MV$, where $v,V$ are velocities of cannon ball and cannon respectively. 2. Work done by the ground on the cannon is given by $W=F×s$ Where $s$ is the distance of the recoil of the cannon and $F$is the resistive force. 3. According to the law of conservation of kinetic energy between the ground and cannon, we get work done by the ground on the cannon is equal to the kinetic energy of the cannon immediately after firing. $W=\frac{1}{2}M{V}^{2}$Step 3. Calculations:Putting the known values in the above formulas one by one, we get$⇒2×v=400×V.......\left(1\right)$$⇒W=100×0.5=50J.......\left(2\right)$All the K.E acquired by the cannon will be dissipated in overcoming the frictional force.Hence $50=\frac{1}{2}M{V}^{2}\phantom{\rule{0ex}{0ex}}{V}^{2}=50×2}{400}\phantom{\rule{0ex}{0ex}}V=0.5m}{s}$Putting the value of $V$ in $\left(1\right)$, we get$v=100m}{s}$Therefore time is taken by the cannonball to hit the target $=\frac{D}{V}$$=\frac{500}{100}=5\mathrm{sec}$.

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