First slide

Conservation of mechanical energy

Question

Two identical balls A and B are released from the position shown in the figure. They collide elastically with each other on the horizontal portion. The ratio of heights attained by A and B after collision is (neglect friction)

Moderate

A

1:4
B

2:1
C

4:13
D

2:5

Solution

When the two balls collide with each other ,as the mass of the two balls is equal, they exchange their velocities on colliding elastically. Let the speed of the ball B when it reaches back to the initial position be v. Then
$t o t a l e n e r g y o f B a f t e r c o l l i s i o n = {(K E + P E)}_{a t p o i n t B} 4 m g h = \frac{1}{2} m v^{2} + m g h \Rightarrow v = \sqrt{6 g h}$
Height reached by particle B(from highest point on the incline) is , ball is expected to have projectile motion from point B
$m a x i m u m h e i g h t r e a c h e d b y b a l l f r o m p o i n t B i s H_{B} = \frac{v^{2} \sin^{2} θ}{2 g} H_{B} = \frac{v^{2} \sin^{2} 60^{0}}{2 g} H_{B} = \frac{(6 g h) {(\frac{\sqrt{3}}{2})}^{2}}{2 g} H_{B} = \frac{9 h}{4}$ ,
total height from the bottom reached by ball B = h+ $\frac{9 h}{4}$ = $\frac{13 h}{4}$
After collision the particle A reaches the maximum height = h
Ratio = $d u e t o e x c h a n g e o f v e l o c i t y, b a l l A w i l l b e a b l e t o r e a c h m a x i m u m h e i g h t h b u t b a l l B r e a c h e s m a x i m u m h e i g h t = \frac{13 h}{4} \frac{H_{A}}{H_{B}} = \frac{h}{\frac{13 h}{4}} \frac{H_{A}}{H_{B}} = \frac{4}{13}$

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