Three identical balls of same mass I, II & III are placed on a smooth floor in a straight line at a separation of 10 m between them as shown. Initially the balls are stationary. Ball I is given velocity of 10m/s towards ball II. Collision between ball I & II is inelastic with coefficient of restitution 0.5 but collision between ball II & III is perfectly elastic. Find the time interval between those two consecutive collision [between ball I and II].

Engineering

Physics

Collision

Question

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This problem involves multiple collisions with different coefficients of restitution. The key is to analyze the velocities after each collision using conservation of momentum and the restitution formula.

For the first collision (I & II, e=0.5):

Initial: $u_{1} = 10 m / s$ , $u_{2} = 0$ .

Momentum: $m v_{1} + m v_{2} = m \times 10$ .

Restitution: $v_{2} - v_{1} = 0.5 \times 10$ .

Solving gives $v_{1} = 2.5 m / s$ , $v_{2} = 7.5 m / s$ .

Ball II then collides elastically (e=1) with stationary Ball III, exchanging velocities: Ball II becomes 0 m/s, Ball III becomes 7.5 m/s.

Ball I (2.5 m/s) now chases the stationary Ball II (10m apart). Relative speed is 2.5 m/s. Time until next collision is distance/relative speed = 10/2.5 = 4 seconds.

Final Answer: 4 s