A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by impact, the value of coefficient of restitution is
A) 1 / (2√2)
B) 1 / √3
C) 1 / √2
D) √3 / 2

To solve this problem, we need to analyze the situation using the concepts of coefficient of restitution and kinetic energy loss during the impact.
Coefficient of Restitution (e)
The coefficient of restitution (ee) between two objects is a measure of how elastic the collision is. It is defined by the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact:
e=v2′−v1′v1−v2e = \frac{{v_2' - v_1'}}{{v_1 - v_2}}
Where:
• v1v_1 and v2v_2 are the initial velocities of the two objects (before the collision),
• v1′v_1' and v2′v_2' are the final velocities of the two objects (after the collision).
In this case:
• The first ball is moving with an initial velocity v1v_1,
• The second ball is at rest initially (v2=0v_2 = 0),
• After the impact, the first ball is brought to rest, so v1′=0v_1' = 0,
• The second ball moves with a final velocity v2′v_2'.
Kinetic Energy Considerations
The problem also states that half of the kinetic energy is lost during the impact. This implies that the final kinetic energy of the system is half of the initial kinetic energy.
• The initial kinetic energy of the system is:
Kinitial=12mv12K_{\text{initial}} = \frac{1}{2} m v_1^2
• After the collision, the final kinetic energy is:
Kfinal=12mv2′2K_{\text{final}} = \frac{1}{2} m v_2'^2
Since half of the kinetic energy is lost, we have:
Kfinal=12Kinitial=12×12mv12=14mv12K_{\text{final}} = \frac{1}{2} K_{\text{initial}} = \frac{1}{2} \times \frac{1}{2} m v_1^2 = \frac{1}{4} m v_1^2
This leads to the equation for the final velocity of the second ball:
12mv2′2=14mv12\frac{1}{2} m v_2'^2 = \frac{1}{4} m v_1^2
Solving for v2′v_2':
v2′2=12v12v_2'^2 = \frac{1}{2} v_1^2 v2′=v12v_2' = \frac{v_1}{\sqrt{2}}
Using the Coefficient of Restitution
Now, applying the coefficient of restitution formula:
e=v2′−v1′v1−v2=v2′−0v1−0=v2′v1e = \frac{{v_2' - v_1'}}{{v_1 - v_2}} = \frac{{v_2' - 0}}{{v_1 - 0}} = \frac{{v_2'}}{{v_1}}
Substitute v2′=v12v_2' = \frac{v_1}{\sqrt{2}} into the equation:
e=v12v1=12e = \frac{{\frac{v_1}{\sqrt{2}}}}{{v_1}} = \frac{1}{\sqrt{2}}
The coefficient of restitution ee is 12\frac{1}{\sqrt{2}}. Therefore, the correct answer is:
12\boxed{\dfrac{1}{\sqrt{2}}}
This corresponds to option C.