Question

In a collinear collision, a particle with an initial speed ${v_0}$ ​strikes a stationary particle of the same mass. If the final total kinetic energy is $50\% $ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is?
(A) $\dfrac{{{v_0}}}{2}$
(B) $\dfrac{{{v_0}}}{{\sqrt 2 }}$
(C) $\dfrac{{{v_0}}}{4}$
(D) $\sqrt 2 {v_0}$

Answer 1

Hint: To solve this question, we need the values of the velocities of the particles after collision. To find the velocities, we can use the conservation of momentum theorem and the kinetic energy formula. The equation of the theorem will give values of velocities and by using that we find the relative velocity between the particles after collision.

Formula Used:
According to Conservation of momentum,
Sum of momentum of particles before collision = Sum of momentum of particles after collision
$ \Rightarrow {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Where ${m_1}$and ${m_2}$are the masses of the particle involved in collision, ${u_1}$ and ${u_2}$ are the velocities of the particles before collision, ${v_1}$ and ${v_2}$ are the velocities of the particles after collision.
Kinetic energy of a body is $\dfrac{1}{2}m{v^2}$ where $m$ is the mass of the body and $v$ is the velocity of the body.

Complete step by step answer:
In the question it’s given that two particles with same mass but one at a velocity of ${v_0}$ and other at rest (zero velocity) undergo collinear collision.
Let the mass of both the particles be $m$. The initial velocity of the particle before collision is ${v_0}$ . Let the velocities of the particles after collision be ${v_1}$ and ${v_2}$.
Applying the conservation of momentum,
$ \Rightarrow {m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}$
Substituting the values of the mass of the particles and the velocities of the particles before and after collision we get,
$ \Rightarrow m{v_0} = m{v_1} + m{v_2}$
$ \Rightarrow {v_0} = {v_1} + {v_2}$
Let this be equation 1.
Also it’s given in the question the kinetic energy of the particles increases by $50\% $ after collision.
Kinetic energy before collision will be,
$K{E_{beforecollision}} = \dfrac{1}{2}m{v_0}^2$
Kinetic energy after collision will be,
$K{E_{aftercollision}} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2$
Using this information,
 $ \Rightarrow \dfrac{3}{2}K{E_{beforecollision}} = K{E_{aftercollision}}$
$ \Rightarrow \dfrac{3}{2} \times \dfrac{1}{2}m{v_0}^2 = \left( {\dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2} \right)$
$ \Rightarrow \dfrac{3}{2}{v_0}^2 = {v_1}^2 + {v_2}^2$
Let this be equation 2.
After squaring both sides of equation 1, we get
$ \Rightarrow {v_0}^2 = {\left( {{v_1} + {v_2}} \right)^2}$
$ \Rightarrow {v_0}^2 = {v_1}^2 + {v_2}^2 + 2{v_1}{v_2}$
Let this be equation 3.
Substituting the value of ${v_1}^2 + {v_2}^2$ from equation 2 in equation 3 we get
$ \Rightarrow {v_0}^2 = \dfrac{3}{2}{v_0}^2 + 2{v_1}{v_2}$
$ \Rightarrow {v_1}{v_2} = - \dfrac{{{v_0}^2}}{4}$
Let this be equation 4.
We know the relative velocity between particle 1 and 2 is
${\vec v_1} - {\vec v_2} = {\vec v_r}$
$ \Rightarrow {v_r} = \sqrt {{{\left( {{v_1} - {v_2}} \right)}^2}} = \sqrt {{v_1}^2 + {v_1}^2 - 2{v_1}{v_2}} $
Substituting the value of ${v_1}^2 + {v_2}^2$ from equation 2 and ${v_1}{v_2}$from equation 4, we get
$ \Rightarrow {v_r} = \sqrt 2 {v_0}$

Hence, option (D) is the correct option.

Note: To solve questions with more than one variable, we need at least two equations. As the velocities of the individual particles were not given in the question, the relative velocity was found using the velocity of the particle before collision. Also we should be careful while using the formulae for calculating kinetic energy.