Question

A bullet of mass $10g$ traveling horizontally with a velocity of $150m/s$ strikes a stationary wooden block and comes to rest in $0.03s$. Calculate the distance of penetration of the bullet into the block. Also, calculate the magnitude of the force exerted by the wooden block on the bullet.

Answer 1

Step 1: Given data

1. The mass of the bullet is $m=10g.$
2. The initial speed of the bullet is $u=150m/s.$
3. The final velocity of the bullet is $v=0.$
4. The bullet will stop at a time $t=0.03sec.$

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 formulas 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.
3. And $v=u+at$.

Step 3: Finding the distance of penetration

Let a be the acceleration of the bullet and s is the distance of penetration of the bullet inside the block.

Now,

$$\begin{gathered} v=u+at \\ ora=\frac{v-u}{t}=\frac{0-150}{0.03}=-5000 \\ ora=-5000m/s^2...................\left(1\right) \end{gathered}$$

Again,

$$\begin{gathered} v^2=u^2+2as \\ ors=\frac{v^2-u^2}{2a}=\frac{0-{\left(150\right)}^2}{2\times \left(-5000\right)}=2.25 \\ ors=2.25 \end{gathered}$$

So, the distance of penetration of the bullet in the wooden block is $2.25m$.

Step 4: Finding the magnitude of the force

$$\begin{gathered} F=ma \\ orF=0.01\times \left(-5000\right)=50 \\ orF=-50N...................\left(2\right) \end{gathered}$$

The negative sign in equation (1) is due to the negative acceleration (retardation) of the bullet inside the block and similarly, in equation (2) the force is a resistive force.

So, the magnitude of force by the wooden block on the bullet is $-50N$.