Question

A bullet of mass $0.04kg$ 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

Hint: Start by calculating the value of acceleration by using the third equation of motion, i.e. ${v^2} = {u^2} + 2as$ . Then calculate the average force exerted on the block by using Newton’s second equation of motion, i.e. $F = ma$ .

Complete answer:
By the application of Newton’s second law of motion, we know that the force applied on any object is given by the equation
$F = ma$
Here, $F = $ Force applied to the object
$m = $ The mass of the object
$a = $ The acceleration of the body
In the problem the mass of the bullet is given, so we only have to calculate the acceleration of the bullet.
Acceleration is a vector quantity that is described as the rate by which a change of velocity occurs with time.
Now, we know that when the bullet will strike the block, the bullet will slowly decelerate until it comes to a stop (so $v = 0$ ).
The third equation of motion is as follows
${v^2} = {u^2} + 2as$
Here, $v = $ The final velocity of an object
$u = $ The initial velocity of an object
$a = $ The acceleration of the object
$s = $ The distance covered by the object
Given in the problem, $u = 90m/s$
$s = 60cm = 0.6m$
So the third equation of motion becomes
${\left( 0 \right)^2} = {\left( {90} \right)^2} + 2 \times a \times 0.6$
$a = - 6750m/{s^2}$
Now the resistive force applied by the block on the bullet can be calculated by using Newton’s second law of motion
$F = ma$
$F = 0.04 \times - 6750$
$F = - 270N$

Note:
Force is equal to mass into acceleration. Here the negative sign indicates that force applied by the rock is in the opposite direction to the velocity of the bullet. Third law of friction, when we touch a heavy box, it pushes back at you with an equal and opposite force so that the harder the force of action, the greater the force of reaction.