Question

A bullet of mass 40 g moving with a speed of $90m{s}^{-1}$ enters a heavy wooden block and is stopped after a distance of 60 cm. The average resistance force exerted by the block on the bullet is

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

Answer 1

The correct option is C 270 N

Here, u=90 $m{s}^{-1}$, v= 0

m=40gm=$\frac{40}{1000}kg$= 0.04 kg, s=60 cm= 0.6 m

Using $v^2-u^2=2as$

$\therefore (0{)}^2-(90{)}^2=2a\times 0.6$

$a=\frac{(90{)}^2}{2\times 0.6}=-6750m{s}^{-2}$

-ve sign shows the retradation.$\therefore$ The average resistive force exerted by block on the bullet is

$F=m\times a=\left(0.04kg\right)\left(6750m{s}^{-2}\right)=270N$