Question

A bullet of mass $40 \,g$ moving with a speed of $90\, m s^{-1}$ 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. $220\, N$
• C. $270 \,N$
• D. $320\, N$

Answer 1

Answer: (C)

$270 \,N$

Answer 2

Here, $u=90 \,m \, s^{-1},$ $v=0$ $m=40 \,g$ $=\frac{40}{1000} kg$ $=0.04\, kg,$ $s=60\, cm$ $=0.6\, m$ using $v^{2}$ $-u^{2}$ =$2 as$ $\therefore$ $\quad\quad\left(0\right)^{2}$ $-\left(90\right)^{2}$ $=2a\times0.6$ $\quad\quad a=-\frac{\left(90\right)^{2}}{2\times0.6}$ $=-6750\, m s^{-2}$ $-ve$ sign shows the retardation. $\quad\therefore\quad$ The average resistive force exerted by block on the bullet is $\quad F=m\times a$ $=\left(0.04 \, kg\right)$ $\left(6750\,ms^{-2}\right)$ $=270 \, N$