Question

A bullet of mass m1 traveling with a velocity v strikes a stationary wooden block of mass m2 and gets embedded into it determine the expression for loss in the kinetic energy of the system is this violating the principal of conservation of energy if not how can you account for this loss
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Answer 1

The Situation refers to Inelastic Collision

Explanation:

The problem does not specifically state this, but I'm going to assume that the bullet becomes embedded in the block of wood when it strikes, making this an "inelastic collision". If that's not the case, then we aren't given enough information to solve the problem.

If the bullet strikes and embeds in the wooden block, the block/bullet combined mass is going to have some velocity - We'll call this $v_f$ (velocity after the collision). The block/bullet then swings from the rope up to a height of "h".

For this motion, the conservation of energy applies:

 

The initial total energy (kinetic + potential) = the final total energy (kinetic + potential)

 

$KE_0 + PE_0 = KE_f + PE_f$

 

   Right after the bullet hits the block, the initial height is 0, so $PE_0=0$. When the bullet/block reaches its maximum height, it is at rest, so $KE_f=0$

 

$KE_0 = PE_f$

 

$(\frac {1}{2})(m+M)v_f^2 = (m+M)gh$

 

We can use this equation to solve for "$v_f$", the velocity of the bullet/block after the collision:

 

 $v_f = \sqrt(2gh)$

Now let's go back in time to look at the collision itself. For an inelastic collision (where the objects become stuck together), the initial momentum = final momentum (just like ANY other collision).  

 

$p_0 = p_f$

 

$m_{bullet} \times v_{0,bullet} + M_{block} \times v_{0,block} = (m_{bullet} + M_{block}) \times v_f$

 

Since the block is not moving before the collision, $v_{0, block} = 0$

 

$m_{bullet} \times v_{0,bullet} = (m+M) \times v_f$

 

Rearranging to solve for in initial velocity of the bullet, $"v_0":$

 

$v_0 = \frac {(m+M) \times v_f}{m}$

 

Substituting in the expression for "$v_f$" that we solved for earlier:

 

  $v_0 = \frac {(m+M)\times \sqrt(2gh)}{m}$