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Question: A bullet of mass m travelling with a speed v hits a block of mass M initially at rest and gets embed...

A bullet of mass m travelling with a speed v hits a block of mass M initially at rest and gets embedded in it. The combined system is free to move and there is no other force acting on the system. The heat generated by the process will be:
(A) Zero
(B) 2mv22m{v^2}​
(C) 12(mMm−M)v2\dfrac{1}{2}\left( {\dfrac{{mM}}{{m - M}}} \right){v^2}
(D) 12(mMm+M)v2\dfrac{1}{2}\left( {\dfrac{{mM}}{{m + M}}} \right){v^2}

Explanation

Solution

Use the Law of linear momentum to solve the sum. Also, take the initial speed of the block as 0 before the bullet hits it.

Complete step by step answer:
First write all the values which have been provided,
Mass of bullet= m1=m{m_1} = m
Initial speed of bullet =u1=v{u_1} = v
Mass of block=m2=M{m_2} = M
Initial speed of block =u2=0{u_2} = 0
Now we can take the common velocity of the bodies after collision =V
So According to conservation of linear momentum,
⇒m×v+M×0=(m+M)V\Rightarrow m \times v + M \times 0 = (m + M)V
⇒V=mv(m+M)\Rightarrow V = \dfrac{{mv}}{{(m + M)}}after simplification
Now we know that the,
Heat generated = loss in Kinetic energy
Which also means that,
Initial Kinetic energy =Final Kinetic Energy
Now writing the equation accordingly,
⇒12mv2=12(m+M)V2\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{2}(m + M){V^2}
Now we know that V=mv(m+M)V = \dfrac{{mv}}{{(m + M)}} so putting the value of v in the above equation we get,
⇒12mv2=12(m+M)(mv(m+M))2\Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{2}(m + M){(\dfrac{{mv}}{{(m + M)}})^2}
By doing further simplification we get,
⇒12mv2−12(m+M)m2v2(m+M)2=0\Rightarrow \dfrac{1}{2}m{v^2} - \dfrac{1}{2}(m + M)\dfrac{{{m^2}{v^2}}}{{{{(m + M)}^2}}} = 0
Now cancelling one (m+M)(m + M)we get,
⇒12mv2−12m2v2(m+M)=0\Rightarrow \dfrac{1}{2}m{v^2} - \dfrac{1}{2}\dfrac{{{m^2}{v^2}}}{{(m + M)}} = 0
After simplification,
⇒  12mv2(1−m(m+M))=0\Rightarrow \;\dfrac{1}{2}m{v^2}\left( {1 - \dfrac{m}{{(m + M)}}} \right) = 0
Now by cancelling m we get,
⇒12mv2(M(m+M))=0\Rightarrow \dfrac{1}{2}m{v^2}\left( {\dfrac{M}{{(m + M)}}} \right) = 0
Now rearranging we get,
⇒12(mMm+M)v2\Rightarrow \dfrac{1}{2}\left( {\dfrac{{mM}}{{m + M}}} \right){v^2}
Therefore, the correct option is d).
Note: Students need to conceptualize the sum before attempting to do it. The loss in Kinetic energy of the bullet will be equal to heat generated in the gun and also in the recoil of the gun.Kinetic energy is the energy possessed by an object in motion. Kinetic energy is directly proportional to the mass of the object and to the square of its velocity.