Question

The average force necessary to stop a bullet of mass 20 g moving with a speed of \[250m{{s}^{-1}}\], as it penetrates into the wood for a distance of 12 cm is
A. \[2.2\times {{10}^{3}}N\]
B. \[3.2\times {{10}^{3}}N\]
C. \[4.2\times {{10}^{3}}N\]
D. \[5.2\times {{10}^{3}}N\]

Answer 1

Hint: Stopping a bullet refers to the fact that its final velocity becomes zero. We can use kinematic equations to solve this problem. Here we have to find the force required to stop the bullet. So this force is a resistive force towards the motion of the bullet.
Formula used:
\[{{v}^{2}}-{{u}^{2}}=2aS\], where v is the final velocity, u is the initial velocity, a is the acceleration and S s the displacement.
\[F=ma\], where m is the mass of the bullet and a is the acceleration of the bullet.

Complete step by step answer:
  According to the kinematic equation,
\[{{v}^{2}}-{{u}^{2}}=2aS\], where v is the final velocity, u is the initial velocity, a is the acceleration and S s the displacement.
Here the bullet is travelling with an initial velocity of \[250m{{s}^{-1}}\]. The stopping of a bullet refers to its final velocity becoming zero. Thus, it can’t move anymore. We don’t have to consider the path of the bullet before the hitting of the bullet on the wood. So the displacement is already given in the question. It is about 12 cm. We have to convert this into the SI unit.
So the displacement is equal to the 0.12 metre.
We can substitute these data into the kinematic equation.
\[-{{250}^{2}}=2a\times 0.12\]
From this, we can find out retardation of the bullet.
\[\dfrac{-{{250}^{2}}}{2\times 0.12}=a\]
\[a=26.04\times {{10}^{4}}m{{s}^{-2}}\]
The acceleration of the bullet will be, \[a=26.04\times {{10}^{4}}m{{s}^{-2}}\]
From the acceleration, we can find out the force acting on the bullet.
\[F=ma\], where m is the mass of the bullet and a is the acceleration of the bullet.
Here, the mass of the bullet is 20 g. It has to be converted into the SI unit.
So the mass becomes .02 kg.
\[\begin{align}
  & F=0.02\times 26.04\times {{10}^{4}} \\
 & =5208N \\
\end{align}\]
So \[5.2kN\] force required to stop the bullet. Hence the correct option is D.
Note: You should remember Newton’s first law that each body continues its state of rest or uniform motion along a straight line, unless it is compelled by an external force. The force acting on the bullet is trying to stop the bullet. When it stops the bullet, the final velocity of the bullet will become zero. This is the key point to find the acceleration of the bullet. Candidates are advised to learn the four kinematic equations. With these kinematic equations, you can solve all the most of the problems related to motion.
\[{{v}^{2}}-{{u}^{2}}=2aS\]
\[\begin{align}
  & v=u+at \\
 & S=ut+\dfrac{1}{2}a{{t}^{2}} \\
 & S=\left[ \dfrac{u+v}{2} \right]t \\
\end{align}\]
where v is the final velocity, u is the initial velocity, t is the time, S is the displacement and a is the acceleration.