A 5.00-g bullet is fired horizontally into a 1.50-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.0250 m along the surface before stopping. What was the initial speed of the bullet?

1

Answers

2016-04-09T15:10:14+05:30
Let the initial speed of the bullet be v.
Therefore, the initial kinetic energy of the bullet = (1/2) X m X (v X v)
Since, the bullet gets stuck to the block after hitting, so, it loses its initial kinetic energy completely and from the principle of energy conservation (the First Law of Thermodynamics), we can say that this is the energy gained by the block. If we neglect the rise in temperature (alternatively, rise in its molecular internal energy) of the block when the bullet hits it, this gain in energy can only be gain in its kinetic energy (mechanical energy).

Therefore, the total mechanical energy of the block after the bullet hits it but before it starts to slide on the horizontal floor  = total mechanical energy of the block before the bullet hits + gain in its mechanical energy after hitting = 0 +  (1/2) X m X (v X v), considering the horizontal surface to be our datum for measurement of height and thus gravitational potential energy.
 
* We may note that the block has zer
gravitational potential energy before the bullet hits it as well as after hitting.

Now, the total friction force between the block and the surface  = Coefficient of kinetic friction X Normal reaction to the block from the floor
Coefficient of kinetic friction X Weight of the block (since there is no movement in the direction normal to the plane, therefore, from the condition of equilibrium of the block in the normal direction, we can conclude that weight of the block = normal reaction to the block from the floor).

That is the total friction force between the block and the surface  =  Coefficient of kinetic friction X Mass of the block X Acceleration due to gravity 

Therefore, the work done against the friction force = T
otal friction force between the block and the surface X Distance through which the block slides 
 Coefficient of kinetic friction X Mass of the block X Acceleration due to gravity X Distance through which the block slides 
= 0.20 X 1.5 kg X 9.81 m/(second squared) X 0.0250 m
= 0.073575 J
Therefore, again from the principle of conservation of energy, we can say that this is the loss in mechanical energy of the block when it stops

Now, the loss in mechanical energy of the block = Initial total mechanical energy - Final total mechanical energy
=  
(1/2) X m X (v X v) - 0
(1/2) X m X (v X v)

Therefore, 
(1/2) X m X (v X v) = 0.073575 J
So, v = Square root of [(
0.073575 X 2)/1.5]  (As m = 1.5 kg)
         = 0.3132 m
         = 31.32 cm.
0