Ball A of mass 0. 5kg moves towards the left at 3 m s^-1 on a smooth track. Then it collides with a stationary ball B of mass 0.4 kg. B moves up the slope and reaches point P(which is 0.3 m above the horizontal track) before it moves down the slope.
Find the speeds of A and B just after the collision.
Physics: force and motion
2012-07-25 10:43 pm
回答 (1)
2012-07-26 1:13 am
✔ 最佳答案
I assume that the slope is smooth.Let v be the speed of ball B just after collision. By conservation of energy,
(1/2).(0.4)v^2 = 0.4g.(0.3)
where g is the acceleration due to gravity (= 10 m/s^2)
hence, v = square-root[2g x 0.3] m/s = 2.449 m/s
By conservation of momentum,
Initial momentum = 0.5 x 3 kg.m/s = 1.5 kg.m/s
Final momentum = (0.5u + 0.4 x 2.449) kg.m/s
where u is the speed of ball A just after collision
Hence, 1.5 = 0.5u + 0.4 x 2.449
u = 1.04 m/s
收錄日期: 2021-04-13 18:51:21
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