Question

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 is
A. 180N
B. 220N
C. 270N
D. 320N

Answer 1

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` .