Please wait...
Contact Us
Need assistance? Contact us on below numbers

For Study plan details

10:00 AM to 7:00 PM IST all days.

For Franchisee Enquiry



Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

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
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
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer 10/10

Your answer has been posted successfully!

Chat with us on WhatsApp