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

Options

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

Answer

C

Explanation

Here, `u=90ms^(-1), v=0`
`m=40g =(40)/(1000)kg=0.04kg, s=60cm=0.6m`
`"Using" v^(2)-u^(2)=2as`
`therefore (0)^(2)-(90)^(2)=2a xx0.6`
`a=-((90)^2)/(2xx0.6)=-6750"ms"^(-2)`
-ve sign shows the retardation.
The average resistive force exerted by block on the bullet is
`F=mxxa,=(0.04kg)(6750 ms^(-2))=270N`