Question

A pendulum consists of a wooden bob of mass m and length ‘l’. A bullet of mass \[{{m}_{1}}\] is fired towards the pendulum with a speed \[{{v}_{1}}\]. The bullet emerges out of the bob with a speed \[{{v}_{1}}\]/3, and the bob just completes motion along a vertical circle. Then \[{{v}_{1}}\]
A- \[\{\dfrac{m}{{{m}_{1}}}\}\sqrt{5gl}\]
B- \[\dfrac{3}{2}\{\dfrac{m}{{{m}_{1}}}\}\sqrt{5gl}\]
C- \[\dfrac{2}{3}\{\dfrac{{{m}_{1}}}{m}\}\sqrt{5gl}\]
D- \[\{\dfrac{{{m}_{1}}}{m}\}\sqrt{gl}\]

Answer 1

Hint: Since, there is no external force acting the total momentum of the system must be conserved that is total momentum before Collision must be equal to the total momentum after collision. Momentum depends upon the variables mass and velocity of the body involved.

Step by step answer:Initially pendulum was at rest and the bullet was moving.
 mass of the bullet, \[{{m}_{1}}\]
mass of the pendulum, m
length of the pendulum, l
Initial velocity of bullet, \[{{v}_{1}}\]
Initial velocity of pendulum= 0 m/s
Final velocity of the bullet = \[{{v}_{1}}\]/3
Final velocity of pendulum = v
Using the law of conservation of the momentum, we get
 \[{{m}_{1}}{{v}_{1}}+0={{m}_{1}}\times \dfrac{{{v}_{1}}}{3}+mv\]
\[
   \Rightarrow {{m}_{1}}{{v}_{1}}={{m}_{1}}\times \dfrac{{{v}_{1}}}{3}+mv \\
  \Rightarrow {{m}_{1}}{{v}_{1}}-\dfrac{{{m}_{1}}{{v}_{1}}}{3}=mv \\
 \Rightarrow \dfrac{2{{m}_{1}}{{v}_{1}}}{3}=mv \\
\]
\[\Rightarrow {{v}_{1}}=\dfrac{3mv}{2{{m}_{1}}}\]------(1)
Now it is given in the question that bob just completes motion along a vertical circle,
\[\because v=\sqrt{5gl}\]
Putting this in eq (1) we get,
\[\therefore \dfrac{3}{2}\{\dfrac{m}{{{m}_{1}}}\}\sqrt{5gl}\]

Hence, the correct option is (B).

Note: This is the easiest way to solve this problem. Momentum conservation is a universal law which holds until there are no external force acts on the system. The most important thing to notice in such questions is what are the factors which affect the position of the bob.