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Question

A lead bullet of mass 2kg, traveling 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.


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Solution

Step 1: Given data

  1. The initial velocity of the bullet, u=20m·s-1.
  2. As the bullet comes to rest thus the final velocity of the bullet, v=0m·s-1.
  3. Displacement before the bullet stops, s=20m.

Step 2: Determine the acceleration of the bullet

  1. Let the acceleration of the bullet be a.
  2. It is known that, v2=u2+2as.

a=v2-u22sa=02-2022·20a=-40040a=-10

Thus, the acceleration is -10m·s-2, negative sign implies that the velocity is decreasing.

Step 3: Determine the resistive force applied by the target

The acceleration of the bullet is a=-10m·s-2.

The mass of the bullet is m=2kg.

Thus, the force applied by the target can be given by, F=ma

F=2×-10NF=-20N

A negative sign indicates that the force is resistive.

Therefore, the resistive force is 20N.

Hence, the resistive force applied by the target is 20N and the acceleration of the bullet is -10m·s-2.


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