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A ball whose mass is 100 g is dropped from a heightof 2 m from the floor. It rebounds vertically up-wards after colliding with the floor to a height 1.5m. Find (a) the momentum of the ball before andafter colliding with the floor, (b) the average forceexerted by the floor on the ball. Assume that colli-sion lasts for 10^- 8 s. |
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Answer» V2−U2 = 2gS Here given that, U = 0 m/s, S = 1.5 m. So,V2−02 = 2×10×2 ⇒V=40−−√ m/s.So, momentum of the ball is,Pi=mV = 0.1×40−−√ = 0.1×6.325 = 0.633⇒ Pi=0.633 kg.m/s After collision, it is given that the ball rises to a height S' = 1.5 m.So, velocity with which the body rises is, V2−U2=−2gS' ⇒ 02 − V'2 = −2×10×1.5 ⇒V' = 30−−√ m/s.So, the momentum of the ball, when the ball is bouncing back. Pf=mV' = 0.1×30−−√⇒Pf=0.547 kg m/s. Therefore, the average force exerted by the floor on the ball is, Favg=∆PDt = 0.633−0.54710−8=0.086×108 ⇒Favg=8.6×10^6N Read more on Brainly.in - https://brainly.in/question/4699679#readmore |
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