Question

A bullet of mass 5 g travelling at a speed of 120 m/s penetrates deeply into a fixed target and is brought to rest in 0.01 s. Calculate the average force exerted on the bullet
(A) 10 N
(B) 20 N
(C) 40 N
(D) 60 N

Answer 1

Hint:To find the average force which is exerted on the bullet by the target we following Newton’s second law of motion need to find the acceleration of the bullet after it enters the target. Since the bullet comes to rest this acceleration will come out to be negative.

Complete step by step answer:
Mass of the bullet,
m= 5g
= \[\dfrac{5}{1000}\]kg
=0.005 kg
The initial speed of the bullet, u= 120m/s
The final speed of the bullet, v= 0 m/s [since it comes to rest]
Time taken, t= 0.01s
To find the acceleration using Newton’s first equation of motion, v=u+at
0=120+0.01a
-120=0.01a
a= -12000 \[m/{{s}^{2}}\]
Now using Newton’s second law,
F=ma
Since this force acts in the opposite direction, it will come out to be negative.
\[\begin{align}
  & F=0.005\times (-12000) \\
 & F=-60N \\
\end{align}\]
Negative sign here signifies the direction, so the resistive force is 60N
Hence, the correct option is (D)

Note:Retarding force is that force which opposes the motion of the body on which it is being acted. Remember force is a vector quantity and so its direction is also to be taken into account. Such problems can be briskly solved using Newton’s equations of motions. Also problems involving bullets can also be based on laws of conservation of momentum.