Question

A bullet of mass $20g$ and moving with a velocity of $200m/s$ strikes a heap of sand and comes to rest after penetrating $3cm$ inside it. The force exerted by the sand on the bullet will be:

• A. $11.2\times {10}^{8}dyne$
• B. $15.7\times {10}^{8}dyne$
• C. $13.3\times {10}^{8}dyne$
• D. $18.6\times {10}^{8}dyne$

Answer 1

The correct option is C $13.3\times {10}^{8}dyne$

Given:

• mass of the bullet (m) = 0.02 kg
  
• initial velocity of the bullet (u) = 200 m/s
  
• final velocity of the bullet (v) = 0 m/s
  
• distance traversed by the bullet in the sand before coming to rest (S) = 0.03 m
  

From the equation of motion,

$v^2–u^2=2aS$

$\Rightarrow 0-{200}^2=2aS$

$\Rightarrow a=-\frac{2}{3}\times {10}^6$ (Here a is the deceleration)

Hence the force exerted by the sand on the bullet is

$F=m(-a)$

$\Rightarrow F=0.02\times \frac{2}{3}\times {10}^6$

$\Rightarrow F=13.3\times {10}^{3}N$

Note: $1N={10}^{5}dyne$

$\Rightarrow F=13.3\times {10}^{8}dyne$

Hence option C is correct.