Question

A bullet of mass $0.04kg$ moving with a speed of $90m/s$ 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

$m=0.04kg$

$u=90m/s$

$v=0m/s$

$s=60cm$$=60\times {10}^{-2}m$

Step 2: Formula used

Newton's third equation of motion $v^2=u^2+2as$

where $u$ is the initial velocity.

$v$ is the final velocity.

$a$ is the acceleration.

$s$ is the displacement.

Newton's second equation of motion $F=ma$

where $F$ is the force.

$m$ is the mass.

Step 3: Calculate the acceleration

Acceleration of bullet, $a=\frac{v^2-u^2}{2s}$

Substitute the given values.

$a=\frac{0-{\left(90\right)}^2}{2\times 60\times {10}^{-2}}$

$=\frac{-90\times 90\times {10}^2}{2\times 60}$

$=-\frac{81}{12}\times {10}^3$

$=-6.75\times {10}^3$

$=-6750m/s^2$

Step 4: Calculate the average force

The average resistive force exerted by block, $F=ma$

Substitute the given values.

$F=0.04\times \left(-6750\right)$

$=-270kgm/s^2$

$=-270N$

Here negative sign indicates that the force applied by the block is in opposite direction to the velocity of the bullet.

Hence, the average force exerted by the block on the bullet is $-270N$.