A body of mass 2kg is projected upward from the surface of the ground at t=0 with a velocity of 20m/s .One second later a body B ,also of mass 2kg is dropped from a height of 20m .If they collide elastically then velocities just after collision are
Asked by m.nilu
22nd September 2018,
2:29 PM
Answered by Expert
Answer:
Let us consider g = 10 m/s2 for the convenience of calculation.
Let t be time taken by the body that is projected upward to meet the another body dropped from 20 m height.
Distance S1 travelled by the body projected upward is given by, S1 = u×t-(1/2)×g×t2 = 20×t -(1/2)×10×t2 = 20×t -5×t2 ..............(1)
Distance S2 travelled by the body dropped from 20 m height is given by, S2 = (1/2)×g×(t-1)2 = 5×(t-1)2 ...............(2)
S1+S2 = 20 = 20×t -5×t2 + 5×(t-1)2 .................(3)
from eqn.(3), we get t = 1.5 s
Let u1 be the velocity of the body projected upward when it meets the body dropped from 20 m height.
u1 = u - g×t = 20 - 10×1.5 = 5 m/s
Let u2 be the velocity of the body which is dropped from 20 m height, when it meets the body that was projected upward.
u2 = g×(t-1) = 10×0.5 = 5 m/s
Total momentum of both body before collision is zero, because bothe the bodies are moving in opposite directions with same speed.
Let v1 be the velocity of body that was projected upward after collision.
Similarly Let v2 be the velocity of body that was dropped from 20 m height after collision.
By conservation of momentum , v1+v2 =0
BY conservation of energy, 
By substituting the values for u1 and u2 , we get v2 = 5 m/s and v1 = -5 m/s
Answered by Expert
23rd September 2018,
3:14 PM
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