Question

A particle \( A \) of mass \( m \) is moving with a velocity \( v_A \), and collides elastically with a particle \( B \), of mass \( 2m \). Particle \( B \) is initially at rest. After collision, \( A \) moves with a velocity \( v_A' \). If \( v_B \) is the final speed of \( B \), then \( v_A'^2 = k v_B^2 \). The value of \( k \) is ...............

Hint:

For elastic collisions, both momentum and kinetic energy are conserved. Use these principles to derive the relationships between the velocities of the colliding particles.

Answer 1

Correct Answer: 1

Step 1: Conservation of momentum.
For an elastic collision, momentum is conserved. The total momentum before and after the collision is equal: \[ m v_A = m v_A' + 2m v_B. \] Simplifying, we get \[ v_A = v_A' + 2v_B. \quad \text{(1)} \] Step 2: Conservation of kinetic energy.
For an elastic collision, the total kinetic energy is also conserved: \[ \frac{1}{2} m v_A^2 = \frac{1}{2} m v_A'^2 + \frac{1}{2} 2m v_B^2. \] Simplifying, we get \[ v_A^2 = v_A'^2 + v_B^2. \quad \text{(2)} \] Step 3: Solve the system of equations.
From equation (1), \[ v_A' = v_A - 2v_B. \] Substitute this into equation (2): \[ v_A^2 = (v_A - 2v_B)^2 + v_B^2. \] Expanding: \[ v_A^2 = v_A^2 - 4 v_A v_B + 4v_B^2 + v_B^2 = v_A^2 - 4 v_A v_B + 5v_B^2. \] Simplifying: \[ 0 = -4 v_A v_B + 5v_B^2 \Rightarrow v_A = \frac{5}{4} v_B. \] Thus, \[ v_A'^2 = \left(\frac{5}{4} v_B - 2v_B\right)^2 = \left(\frac{-3}{4} v_B\right)^2 = \frac{9}{16} v_B^2. \] Step 4: Final result.
Therefore, \( k = \frac{9}{16} \). Final Answer: The value of \( k \) is \( \boxed{\frac{9}{16}} \).