Solveeit Logo
Login

Question

Physics Question on work, energy and power

A bullet of mass mm travelling with a speed vv hits a block of mass MM 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 in the process will be

A

ZeroZero

B

mv22\frac{mv^{2}}{2}

C

Mmv22(M−m)\frac{Mmv^{2}}{2\left(M-m\right)}

D

mMv22(M+m)\frac{mMv^{2}}{2\left(M+m\right)}

Answer

mMv22(M+m)\frac{mMv^{2}}{2\left(M+m\right)}

Explanation

Solution

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
Let the common velocity of the bodies after collision =V=V
According to conservation of linear momentum
m×v+M×0=(m+M)Vm \times v+M \times 0=(m+M) V
V=mv(m+M)V=\frac{m v}{(m+M)}
∴\therefore Heat generated == loss in KEKE
== Initial KE - Final KE =12mv2−12(m+M)v2=\frac{1}{2} m v^{2}-\frac{1}{2}(m+M) v^{2}
=12mv2−12(m+M)(mvm+M)2=\frac{1}{2} m v^{2}-\frac{1}{2}(m+M)\left(\frac{m v}{m+M}\right)^{2}
=12mv2−12m2v2(m+M)=\frac{1}{2} m v^{2}-\frac{1}{2} \frac{m^{2} v^{2}}{(m+M)}
=12mv2(1−mm+M)=\frac{1}{2} m v^{2}\left(1-\frac{m}{m+M}\right)
=12mv2(m+M−mm+M)=\frac{1}{2} m v^{2}\left(\frac{m+M-m}{m+M}\right)
=12mM(m+M)v2=\frac{1}{2} \frac{m M}{(m+M)} v^{2}