Question

What is the value of $ {k_1} $ ?
$ \left( A \right)1 \\
  \left( B \right)2 \\
  \left( C \right)3 \\
  \left( D \right)None{\text{ }}of{\text{ }}these \\ $

Answer 1

Hint :In order to solve this question, we are going to use the concept of the conservation of energy and momentum during the collision, it gives us two equations for the solution of $ {k_1} $ and $ {k_2} $ after solving the two equations by using the techniques of algebra for solving the quadratic equations, and the question is solved.
The conservation of momentum says
 $ {m_1}{v_1} + {m_2}{v_2} = Mv $
And according to the conservation of energy
 $ \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2 = \dfrac{1}{2}M{v^2} $ .

Complete Step By Step Answer:
According to the law of conservation of momentum, the total momentum of the particles before and after the event remains the same.
For the particle $ 1 $
Mass, $ {m_1} = \dfrac{M}{2} $ and the velocity $ {v_1} = {k_1}v $
For the particle $ 2 $
Mass, $ {m_2} = \dfrac{M}{6} $ and the velocity $ {v_2} = {k_2}v $
After the collision the particle that is formed
Mass, $ m = M $ and the velocity $ v = v $
Applying the conservation of momentum
$ {m_1}{v_1} + {m_2}{v_2} = Mv \\
   \Rightarrow \dfrac{M}{2}{k_1}v - \dfrac{M}{6}{k_2}v = Mv \\
   \Rightarrow \dfrac{{{k_1}}}{2} - \dfrac{{{k_2}}}{6} = 1 - - - \left( 1 \right) \\ $
Now, according to the law of conservation of energy, the total kinetic energies of the particles just before the collision is equal to the total kinetic energies of the particles after the collision.
Applying conservation of energy
$ \dfrac{1}{2}{m_1}{v_1}^2 + \dfrac{1}{2}{m_2}{v_2}^2 = \dfrac{1}{2}M{v^2} \\
   \Rightarrow \dfrac{1}{2}M{v^2} + \dfrac{5}{2}M{v^2} = \dfrac{1}{2}\dfrac{M}{2}{\left( {{k_1}v} \right)^2} + \dfrac{1}{2}\dfrac{M}{6}{\left( {{k_2}v} \right)^2} \\
   \Rightarrow 3 = \dfrac{{{k_1}^2}}{4} + \dfrac{{{k_2}^2}}{{12}} - - - - \left( 2 \right) \\ $
Solving $ \left( 1 \right) $ and $ \left( 2 \right) $ by squaring $ \left( 1 \right) $ and subtracting from $ \left( 2 \right) $
We get the equation
$ 3 - 3 = \dfrac{{{k_1}^2}}{4} + \dfrac{{{k_2}^2}}{{12}} - \dfrac{{3{k_1}^2}}{4} - \dfrac{{{k_2}^2}}{{12}} + \dfrac{{{k_1}{k_2}}}{2} \\
  0 = - \dfrac{{{k_1}^2}}{2} + \dfrac{{{k_1}{k_2}}}{2} \\
   \Rightarrow \dfrac{{{k_2}}}{2} = \dfrac{{{k_1}}}{2} \\
   \Rightarrow {k_1} = {k_2} \\ $
Using this in $ \left( 1 \right) $
We get
$ \Rightarrow \dfrac{{{k_1}}}{2} - \dfrac{{{k_1}}}{6} = 1 \\
   \Rightarrow 2{k_1} = 6 \\
   \Rightarrow {k_1} = 3 \\ $
Hence, option $ \left( C \right) $ is correct.

Note :
The two particles of the masses $ {m_1} $ and $ {m_2} $ collide and the resulting mass is $ M $ , the energy and the momentum are conserved . Note that to solve the equations carefully and the two equations are enough for solving the two constants, the constants come out to be equal and putting that in one of the equations gives us the value $ {k_1} = 3 $ .