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 $50 \%$ grater 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 1

The Correct Option is B

To solve this problem, we need to analyze the collision between two particles of equal mass. Let's go through the solution step-by-step:

1. Prior to the collision, let the initial velocity of the moving particle be $v_0$. The stationary particle has an initial velocity of $0$.

2. The initial kinetic energy of the system is solely due to the moving particle. Thus,

   Initial \ Kinetic \ Energy = \frac{1}{2}mv_0^2

3. After the collision, let the velocities of the two particles be $v_1$ and $v_2$, respectively.

   The final kinetic energy is given to be 50\% greater than the initial kinetic energy. Therefore,

   Final \ Kinetic \ Energy = 1.5 \times \frac{1}{2}mv_0^2 = \frac{3}{4}mv_0^2

   The expression for the final kinetic energy is:

   \frac{1}{2}m(v_1^2 + v_2^2) = \frac{3}{4}mv_0^2

   Simplifying the equation:

   v_1^2 + v_2^2 = \frac{3}{2}v_0^2

4. By the principle of conservation of momentum, the total momentum before and after the collision should remain constant. Therefore,

   mv_0 = mv_1 + mv_2 \\ v_0 = v_1 + v_2

5. We have two equations:

   • v_1 + v_2 = v_0
   • v_1^2 + v_2^2 = \frac{3}{2}v_0^2

   To find the magnitude of the relative velocity between the two particles after the collision, calculate:

   (v_1 - v_2)^2 = v_1^2 + v_2^2 - 2v_1v_2

   Using the first equation, v_1 = v_0 - v_2 , substitute in the second equation:

   (\frac{1}{2}v_0^2) + 2(v_0 - v_2)v_2 = (v_1 - v_2)^2 + \frac{3}{2}v_0^2 \\ \Rightarrow (v_1 - v_2)^2 = \frac{3}{2}v_0^2 - (v_1 - v_2)^2

   Solving this gives:

   v_1 - v_2 = \sqrt{4v_0^2 - \frac{3}{2}v_0^2} \\ v_1 - v_2 = \sqrt{2}v_0

6. The magnitude of the relative velocity between the two particles after the collision is therefore \sqrt{2}v_0.

Hence, the correct answer is \sqrt{2}v_0.