Question

A bullet of mass 0.016 kg traveling horizontally at a speed of 280 m/s embeds itself in a block of mass 3 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? vf= m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans, i= J Ktrans, f= J (c) Compare the two results and explain why there is a difference. The internal energy of the block-bullet system has increased. The Energy Principle isn't valid for an inelastic collision. Some of the momentum is lost in an inelastic collision.

Answer 1

(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

$p_i = p_f\\m u_b + M u_B = (m+M)v$

where

m = 0.016 kg is the mass of the bullet

$u_b = 280 m/s$ is the initial velocity of the bullet

M = 3 kg is the mass of the block

$u_B = 0$ is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

$v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s$

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

$K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J$

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

$K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J$

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).