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.
$$\frac{{{v_0}}}{4}$$
B.
$$\sqrt 2 {v_0}$$
C.
$$\frac{{{v_0}}}{2}$$
D.
$$\frac{{{v_0}}}{{\sqrt 2 }}$$
Answer :
$$\sqrt 2 {v_0}$$
Solution :
$$\eqalign{ & \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{2}\left( {\frac{1}{2}mv_0^2} \right) \cr & \Rightarrow v_1^2 + v_2^2 = \frac{3}{2}v_0^2\,......\left( {\text{i}} \right) \cr} $$
From momentum conservation
$$m{v_0} = m\left( {{v_1} + {v_2}} \right)\,......\left( {{\text{ii}}} \right)$$
Squarring both sides,
$$\eqalign{ & {\left( {{v_1} + {v_2}} \right)^2} = v_0^2 \cr & \Rightarrow v_1^2 + v_2^2 + 2{v_1}{v_2} = v_0^2 \cr & 2{v_1}{v_2} = - \frac{{v_0^2}}{2} \cr & {\left( {{v_1} - {v_2}} \right)^2} = v_1^2 + v_2^2 - 2{v_1}{v_2} = \frac{3}{2}v_0^2 + \frac{{v_0^2}}{2} \cr} $$
Solving we get relative velocity between the two particles
$${v_1} - {v_2} = \sqrt 2 {v_0}$$
$$\eqalign{ & \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{2}\left( {\frac{1}{2}mv_0^2} \right) \cr & \Rightarrow v_1^2 + v_2^2 = \frac{3}{2}v_0^2\,......\left( {\text{i}} \right) \cr} $$
From momentum conservation
$$m{v_0} = m\left( {{v_1} + {v_2}} \right)\,......\left( {{\text{ii}}} \right)$$
Squarring both sides,
$$\eqalign{ & {\left( {{v_1} + {v_2}} \right)^2} = v_0^2 \cr & \Rightarrow v_1^2 + v_2^2 + 2{v_1}{v_2} = v_0^2 \cr & 2{v_1}{v_2} = - \frac{{v_0^2}}{2} \cr & {\left( {{v_1} - {v_2}} \right)^2} = v_1^2 + v_2^2 - 2{v_1}{v_2} = \frac{3}{2}v_0^2 + \frac{{v_0^2}}{2} \cr} $$
Solving we get relative velocity between the two particles
$${v_1} - {v_2} = \sqrt 2 {v_0}$$
