Question

A 6.000 g bullet Is fired with a velocity of 600.0 m/s towards a 2.000 kg stationary target. The target has frictionless wheels and rest on a horizontal surface. When the bullet hits the target, it becomes lodged in it, and the target and bullet move forward at a common velocity. Find the velocity of the bullet and target combination after the collision.

Answer 1

This is a good problem on the topic conservation of momentum. In these type of questions students often try to use the energy conservation law. A normal student tries to equate the initial Kinetic energy of the bullet to the final Kinetic energy of the system which is wrong because when the bullet penetrates the body there are some losses due to friction, heat and deformation which prohibits the use of energy conservation law. So in this question we are going to use the conservation of momentum.

Initial momentum of the suystem is:

P_i = (mass of bullet * initial velocity of bullet ) + (mass of stationary targert * initial velocity of target)

P_i = (0.006 kg * 600 m/s) + (2.000kg * 0 m/s)

P_i = (3.6 kg m /sec) +0

so :

P_i = 3.6 kgm/s

Now as the bullet penetrates the target and moves with same speed, hence the final speed of both bullet and target is equal.

let the final velocity of both be v.

So:

P_f = (0.006kg *v m/s)+(2kg * v m/s)

P_f = (2.006 kg * v m/s)

now applying momentum conservation:

P_f = P_i

2.006 * v = 3.6 kg m/s

v = 3.6/2.006 = 1.7946 m/sec