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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 m = 2kg

Mass of cannon M = 400kg

Distance of target D = 500m

Recoiling of cannon s= 50cm

Resistance offered by ground to cannon F = 100N

Step 2. Formula used:

  1. 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 Fis 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=12MV2

Step 3. Calculations:

Putting the known values in the above formulas one by one, we get

2×v=400×V.......1

W=100×0.5=50J.......2

All the K.E acquired by the cannon will be dissipated in overcoming the frictional force.

Hence 50=12MV2V2=50×2400V=0.5ms

Putting the value of V in 1, we get

v=100ms

Therefore time is taken by the cannonball to hit the target =DV=500100=5sec.


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