A ball A of mass 3 m is placed at a distance d from the wall on a smooth horizontal…
Physics · JEE Advanced · NTA Exams — Laws of Motion
A ball A of mass 3m is placed at a distance d from the wall on a smooth horizontal surface. Another ball B of mass m moving with velocity u collides with ball A. The coefficient of restitution between the balls and the wall and between the balls is e :
- the velocity of ball B after collision is \(\frac{u(3 e-1)}{4} .\)
- the velocity of ball B after collision is \(\frac{u(2 e+1)}{4} .\)
- After collision, ball A will move away by distance \(\frac{\mathrm{d}(2 \mathrm{e}+1)}{\mathrm{d}(2 \mathrm{e}-1)} .\)
- After collision, ball A will move away by distance \(\frac{\mathrm{d}(\mathrm{e}+1)}{(3 \mathrm{e}-1)} .\)
Answer
(A) the velocity of ball B after collision is u(3 e-1) 4 ., (D) After collision, ball A will move away by distance d ( e +1) (3 e -1) .
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