Dear Student, Please find below the solution to the asked query: If the balls are released from h height, the velocity gained by both balls just before colliding the ground are, u1=u2=2gh Given that the first ball goes to height h/4 after colliding and the second ball to a height h/16 So, the velocities with which those two bounce back are, v1=2gh4=2gh2 and v2=2gh16=2gh4Therefore, the impulse on both the objects are,I1=∆P1=mu1-v1=m2gh+2gh2=3m2gh2 and I2=∆P2=mu2-v2=m2gh+2gh4=5m2gh4Therefore,I2=56I1⇒5I1=6I2 Hope this information will clear your doubts about the topic If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible Regards

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Dear Student,

Please find below the solution to the asked query:

If the balls are released from h height, the velocity gained by both balls just before colliding the ground are,

$u_{1} = u_{2} = \sqrt{2 g h}$
Given that the first ball goes to height h/4 after colliding and the second ball to a height h/16. So, the velocities with which those two bounce back are,

$v_{1} = \sqrt{2 g \frac{h}{4}} = \frac{\sqrt{2 g h}}{2} a n d v_{2} = \sqrt{2 g \frac{h}{16}} = \frac{\sqrt{2 g h}}{4} T h e r e f o r e, t h e i m p u l s e o n b o t h t h e o b j e c t s a r e, I_{1} = ∆ P_{1} = m (u_{1} - v_{1}) = m (\sqrt{2 g h} + \frac{\sqrt{2 g h}}{2}) = \frac{3 m \sqrt{2 g h}}{2} a n d I_{2} = ∆ P_{2} = m (u_{2} - v_{2}) = m (\sqrt{2 g h} + \frac{\sqrt{2 g h}}{4}) = \frac{5 m \sqrt{2 g h}}{4} T h e r e f o r e, I_{2} = \frac{5}{6} I_{1} \Rightarrow 5 I_{1} = 6 I_{2}$

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

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