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A bullet of mass 40 g moving with a spee...

A bullet of mass `40` g moving with a speed of `90 ms ^(-1)` enters a heavy wooden block and is stopped after a distance of `60` cm . The average resistive force exerted by the block on the bullet is :

A

180 N

B

20 N

C

270 N

D

320 N

Text Solution

Verified by Experts

The correct Answer is:
C
here `u=90" m/s",v=0`
`m=40g=40//1000" kg "=0.04" kg"`
`s=60" cm"=0.6" m"`
Using `v^(2)-u^(2)=2" as"`
Substituting all the given values we get
`a=-6750" m/s"^(2)`, negative sign shows its retardation
So the average force `=0.04xx6750=270N`
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