Question

A bullet of mass 40 g moving with a speed of $90ms^{- 1}$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 $u = 90ms^{- 1},v = 0$

$m = 40g = \frac{40}{1000}kg = 0.04kg$

$s = 60cm = 0.6m$

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

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

-ve sing shows the retardation

$\therefore$The average resistive force exerted by block on the bullet is

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