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Question

A 3 kg steel ball strikes a wall with a speed of 10 ms1 at an angle of 60 with the surface of the wall. The ball bounce off with the same speed and same angle. If the ball in contact with the wall for 0.25 s. The average force exerted by the wall on the ball is

A
753 N
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B
303 N
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C
1503 N
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D
603 N
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Solution

The correct option is C 1503 N



From impulse-momentum theorem,

ΔP=FΔt

Change in momentum of the ball=Impulse on the ball

PfPi=FΔt

[mvcos60^jmvsin60^i][mvcos60^j+mvsin60^i]=FΔt

2mvsin60^i=FΔt

|2mvsin60|=FΔt

F=2mvsin60Δt

Given,

m=3 kg ; v=10 ms1θ=60 ;Δt=0.25 s

F=2×3×10×320.2=1503 N

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (C) is the correct answer.
Why this question ?

Caution: Average force is calculated as ratio of impulse to time of collision.


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