Question

A ball of mass 1.2 kg moving with a velocity of 12 ms$^{-1}$ makes a one-dimensional collision with another stationary ball of mass 1.2 kg. If the coefficient of restitution is $\frac{1}{\sqrt{2}}$, then the ratio of the total kinetic energy of the balls after the collision to the initial kinetic energy is

• A. $\frac{3}{4}$
• B. 1:1
• C. 2:3
• D. 3:5 \bigskip

Hint:

In a one-dimensional collision, the coefficient of restitution (e) determines how much velocity is retained after impact. Using momentum conservation and restitution formulas, we can calculate the final velocities and then find the kinetic energy ratio.

Answer 1

The Correct Option is A

We are given: - Mass of both balls: $m_1 = m_2 = 1.2$ kg. - Initial velocity of first ball: $u_1 = 12$ ms$^{-1}$. - Initial velocity of second ball: $u_2 = 0$ (stationary). - Coefficient of restitution: $$e = \frac{1}{\sqrt{2}}.$$ We need to find the ratio of total kinetic energy after collision to initial kinetic energy. --- Step 1: Apply Velocity Formula after Collision Using the velocity equations for elastic collisions: $$v_1 = \frac{(m_1 - e m_2) u_1 + m_2 (1 + e) u_2}{m_1 + m_2}$$ $$v_2 = \frac{(m_2 - e m_1) u_2 + m_1 (1 + e) u_1}{m_1 + m_2}.$$ Since $m_1 = m_2$, simplifying: $$v_1 = \frac{(1 - e) u_1}{2},$$ $$v_2 = \frac{(1 + e) u_1}{2}.$$ Substituting $e = \frac{1}{\sqrt{2}}$: $$v_1 = \frac{\left(1 - \frac{1}{\sqrt{2}}\right) 12}{2},$$ $$v_2 = \frac{\left(1 + \frac{1}{\sqrt{2}}\right) 12}{2}.$$ Approximating $\frac{1}{\sqrt{2}} \approx 0.707$: $$v_1 = \frac{(1 - 0.707) \times 12}{2} = \frac{(0.293) \times 12}{2} = \frac{3.516}{2} \approx 1.758 \text{ m/s}.$$ $$v_2 = \frac{(1 + 0.707) \times 12}{2} = \frac{(1.707) \times 12}{2} = \frac{20.484}{2} \approx 10.242 \text{ m/s}.$$ --- Step 2: Compute Initial and Final Kinetic Energies # Initial Kinetic Energy: $$KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 = \frac{1}{2} \times 1.2 \times (12)^2.$$ $$= \frac{1.2 \times 144}{2} = \frac{172.8}{2} = 86.4 \text{ J}.$$ # Final Kinetic Energy: $$KE_{\text{final}} = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2.$$ $$= \frac{1}{2} \times 1.2 \times (1.758)^2 + \frac{1}{2} \times 1.2 \times (10.242)^2.$$ $$= \frac{1.2}{2} \times (3.09 + 104.91).$$ $$= 0.6 \times 108 = 64.8 \text{ J}.$$ --- Step 3: Compute the Ratio $$\frac{KE_{\text{final}}}{KE_{\text{initial}}} = \frac{64.8}{86.4} = \frac{3}{4}.$$ Thus, the correct answer is: $$\boxed{\frac{3}{4}}$$ which matches option (1).