Question

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

• A. Acceleration = 20 m/s² and F = -40 N
• B. Acceleration = 40 m/s² and F = +80 N
• C. Acceleration = 40 m/s² and F = -80 N
• D. Acceleration and force are 0

Answer 1

The correct option is B

Acceleration = 40 m/s² and F = +80 N

Given, initial velocity u = 20 $m{s}^{-1}$, final velocity v = 0 $m{s}^{-1}$, distance traveled s = 20 m and mass of bullet = 2kg

We know that, $v^2$​ = $u^2$​ – 2as

0 = ${20}^2-2\times a\times 20$​

On solving,

Acceleration a= 10 $m{s}^{-2}$

So, F = - ma (resistive force) = -$2\times 10$

F = - 20 N