A cricket ball of mass 150 g moving with a speed of 126 kmh-1 hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for 0.001 s, the force that the batsman had to apply to hold the bat firmly at its place would be
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a
10.5 N
b
21 N
c
1.05 × 104N
d
2.1 × 104N
answer is C.
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Detailed Solution
Here, u=-v=126 kmh-1=126×518=35 m/sChange in momentum of the ball, Δp=m(v-u)=1501000(-35-35) =320(-70)=-212 kg-ms-1Now, force, F=ΔpΔt=-21/20.001 N=-1.05×104 NHere, negative sign shows that direction of force will be opposite to the direction of movement of the ball beforehitting.