Question

A lead bullet of mass $ 2kg $ , travelling with a velocity of $ 20m.{s^{ - 1}} $ , comes to rest after penetrating $ 20m $ in a still target. Find the resistive force applied by the target and acceleration of the bullet.

Answer 1

Hint: In order to answer this question, first we will rewrite the given facts of the question, and then we will apply the formula in terms of velocity, acceleration and the distance travelled, and after getting the value of acceleration, we will apply the formula of force in terms of mass and acceleration.

Complete Step By Step Answer:
Given that-
Mass of lead bullet, $ m = 2kg $
Bullet is travelling with a velocity of $ 20m.{s^{ - 1}} $ and then comes to the rest:
So, the initial velocity is, $ u = 20m.{s^{ - 1}} $ .
And the final velocity is, $ v = 0m.{s^{ - 1}} $
and the distance penetrated, $ s = 20cm $ .
So, now we will apply the formula in terms of velocity, acceleration and the distance travelled:
 $ \therefore {v^2} = {u^2} - 2as $
where, $ v $ is the final velocity of the lead bullet,
 $ u $ is the initial velocity of the lead bullet,
 $ a $ is the acceleration.
 $ \Rightarrow {0^2} = {20^2} - 2a.20 \\
   \Rightarrow 40a = 400 \\
   \Rightarrow = a = 10m.{s^{ - 2}} $
Now, to find the force applied by the target, we will apply the formula of force in terms of mass and acceleration:
 $ \therefore F = - ma $
where, $ F $ is the resistive force,
 $ m $ is the mass of a lead bullet,
 $ a $ is the acceleration.
 $ \Rightarrow F = - 2 \times 10 \\
  \therefore F = - 20N $
Hence, the resistive force applied by the target is $ 20N $ .

Note:
A positive value in the formula indicates that the force is acting radially outward (repulsive), whereas a negative value indicates that the force is acting radially inward (repulsive) (attractive). However, because potential energy is a scalar, this convention doesn't matter; you'll get the same result either way.