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Home / Doubts and Solutions/JEE/Physics

A body A moving in a straight line with velocity v makes a collision with a body B initially at rest.After collision B acquires a velocity of 1.6v .Assuming the bodies to be perfectly elastic ,what is the ratio of the mass of A to that of B ?what percentage of A's energy is transferred to B as a result of collision.

Asked by m.nilu 21st September 2018, 6:01 PM
Answered by Expert
Answer:
Let mass of body-A be M and mass of body-B be m. Let w be the velocity of A after collision.
 
By momentum conservation, we have,  M×v = M×w + m×1.6v  or w = v×[ 1 - α×1.6 ] ........................(1)
where α = m/M
 
By Energy conservation, we have, (1/2)×M×v2  = (1/2)×M×w2 + (1/2)×m×2.56×v2 .....................(2)
 
from (2), we write, w2 = (1-2.56×α)×v2 .........................(3)
 
From eqn.(1) and (3), solving for α, we get α = m/M = 1/4 
 
Hence the ratio of mass of body-A to the body-B is 4
 
By substituting α in eqn.(1), we get w = 0.6×v
 
loss of kinetie energy for body-A = (1/2)×M×v2 - (1/2)×M×(0.6×v)2 = (1/2)×0.64×M×v2
 
This loss of energy is transferred to body-B, hence  energy transfered to body-B by body-A is 64% of initial energy of body-A
 
Answered by Expert 22nd September 2018, 2:28 PM
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