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A ball of mass 1kg is dropped from 20m height on ground and it rebounds to height 5m. Find magnitude of change in momentum during its collision with the ground
To solve, the problemfirst, of alluse the 3rd equation to find the value of v (where put the value of u=o, s=20m, g=9.8ms-2) you will get the value of vafter that, again use the third equation to find the u while rebounding (where put the value of v=0 ,s=5m,g=9.8m/s2) you will get the value of u thereafter you must be knowing that, Impulse = change in momentumthen put , mv-(mu) =mv+mu=m(v+u) then put the values of m,v,u you will get the magnitude and change in momentum......
for downward motion ;u=0 m/sa=10 m/s2s=20 musing v2- u2=2asv= 20m/sfor upward motion;v=0 m/sa=10 m/s2s=5 m using v2 – u2=2asu= -10 m/sI=mv – muI=20+10=30 kg m/s
1stly consider coefficient of restitution e^1/2 then apply v^2=2gh =2×10×20 =400 V=20m/s Then v`/v=e^1/2 If v =20 then v`=10 so V`+v=20+10=30m/sNow , p=mv =30×1=30 kg m/s
Dear StudentA ball of mass 1 kg dropped from 20 m height on ground and it rebounds to height 5.Using energy conservation theorem:mgh = ½(mv2)v = √2ghwhere, g = 10 m/s² and h = 20 mv = √2*10*20v = 20 m/secMomentum before collision = mv = 1 x 20 = 20 kgm/sAfter rebounding the ball.Using energy conservation theorem:mgh = ½(mv2)v = √2ghwhere, g = 10 m/s² and h = 5 mv = – √2*10*5v = – 10 m/secMomentum after collision = mv = -1 x 10 = -10 kgm/sChange in momentum during collision = After collision - Before collision = 20 - (-10) = 30 kgm/secHence, The change in momentum is 30 kgm/secI hope this answer will help you.Thanks & RegardsYash Chourasiya
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