Question

A bullet of mass $0.04kg$ moving with a speed of 90 m/s enters a heavy wooden block and is stopped after a distance of $60cm$. What is the average force exerted by the block on the bullet?

Answer 1

Step 1: Given data

1. The mass of the bullet is $m=0.04kg.$
2. The initial speed of the bullet is $u=90m/s.$
3. The final velocity of the bullet is $v=0.$
4. The bullet will stop at a distance is $s=60cm.$

Step 2: Formulae and concept

1. Force is the product of mass and acceleration, i.e, $F=ma$, where, m is the mass of the body and a is the acceleration.
2. From the formulae of kinematics we know, $v^2=u^2+2as$, where, v and u are the final and initial velocities of the bullet s and a are the displacements and a is the acceleration of the body.

Step 3: Finding the force

Let a be the acceleration of the bullet.

Now,

$$\begin{gathered} v^2=u^2+2as \\ ora=\frac{-u^2}{2s}=\frac{-{\left(90\right)}^2}{2\times 0.6}=-6750 \\ ora=-6750m/s^2. \end{gathered}$$

Again,

$$\begin{gathered} F=ma \\ orF=0.04\times \left(-6750\right)=-270 \\ orF=-270N \end{gathered}$$

The negative sign is due to the negative acceleration (retardation) of the bullet inside the block and similarly, the force is a resistive force.

So, the average force by the wooden block on the bullet is $-270N$.