Question

A bullet going with a speed of $350{\text{m}}{{\text{s}}^{ - 1}}$ enters a concrete wall and penetrates a distance of $5{\text{cm}}$ before coming to rest. Find the deceleration.

Answer 1

As the bullet penetrates the wall and comes to rest its velocity decreases. The final velocity of the bullet will be zero. The bullet is said to decelerate as it enters the concrete wall. Newton’s third equation of motion can be utilised to find the deceleration of the bullet.

Formula used:
-Newton’s third equation of motion is given by, ${v^2} - {u^2} = 2as$ where $v$ is the final velocity of the object, $u$ is its initial velocity, $a$ is its acceleration and $s$ is the distance covered by the object.

Complete step by step answer.
Step 1: List the parameters known from the question.
The initial velocity of the bullet is $u = 350{\text{m}}{{\text{s}}^{ - 1}}$ . The distance covered after it enters the wall is $s = 5{\text{cm}}$ .
Since the bullet comes to rest after it penetrates the concrete wall, the final velocity of the bullet must be zero i.e., $v = 0{\text{m}}{{\text{s}}^{ - 1}}$ .
Step 2: Using Newton’s third equation of motion, find the deceleration of the bullet.
Newton’s third equation of motion is given by, ${v^2} - {u^2} = 2as$ -------- (1)
where $v$ is the final velocity of the bullet, $u$ is its initial velocity, $a$ is its acceleration and $s$ is the distance covered by the bullet after it enters the concrete wall.
Equation (1) can be rearranged to express the acceleration of the bullet as $a = \dfrac{{{v^2} - {u^2}}}{{2s}}$ ------ (2)
Substituting values for $u = 350{\text{m}}{{\text{s}}^{ - 1}}$ , $v = 0{\text{m}}{{\text{s}}^{ - 1}}$ and $s = 5{\text{cm}}$ in equation (2) we get, $a = \dfrac{{{0^2} - {{350}^2}}}{{2 \times 0.05}} = - 1.225 \times {10^6}{\text{m}}{{\text{s}}^{ - 2}}$
$\therefore$ the deceleration of the bullet is $- a = 1.225 \times {10^6}{\text{m}}{{\text{s}}^{ - 2}}$

Note: Deceleration refers to the rate at which the bullet slows down. Since it is the opposite of acceleration, deceleration will be negative and can be denoted as $- a$ . When substituting the values for final velocity, initial velocity and distance covered in equation (2), make sure that all these values are expressed in their respective S.I. units. If not, the necessary conversion of units must be done. Here, the distance covered $s = 5{\text{cm}}$ is converted into $s = 0.05{\text{m}}$ while substituting.