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A bullet of mass 40g moving with a speed of `90ms^(-1)` enters a heavy wooden block and is stopped after a direction of 60cm. The average resistive force exered by the block on the bullet isA. 180NB. 220NC. 270ND. 320N |
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Answer» Correct Answer - C Here, `u=90ms^(-1)` , `v=0` . `m=40g(40)/(1000)kg=0.04kg` . `s=60cm=0.6m` . Using `v^(2)-u^(2)=2as` . :. `(0)^(2)-(90)^(2)=2axx0.6` . `:. A=(90)^(2)/(2xx0.6)=-6750ms^(-2)` . `-ve` sign shows the retardation. The average resistive force exerted by block on the bullet is. `F=mxxa=(0.04kg)(6750ms^(-2))=270N` . |
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