Question

Two trolleys X and Y with momenta 20 Ns and 12 Ns respectively travel along a straight line in opposite directions before collision. After collision the directions of motion of both trolleys are reversed and the magnitude of the momentum of X is 2 Ns. What is the magnitude of the corresponding momentum of Y?

Answer 1

answer is 10/3Ns

Answer 2

The goal of open study is to learn...you should post how you got that solution.

Answer 3

i need detailed solution pls

Answer 4

Since this is an elastic collision in one dimension momentum and kinetic energy (KE) will be conserved. Therefore, the sum of the initial and final values will be equal:
\[\large \rho = m_{1i}V_{1i}+m_{2i}V_{2i}=m_{1f}V_{1f}+m_{2f}V_{2f}\]
and\[\large KE=\frac{1}{2}m_{1i}v_{1i}^2+\frac{1}{2}m_{2i}v_{2i}^2=\frac{1}{2}m_{1f}v_{1f}^2+\frac{1}{2}m_{2f}v_{2f}^2\]Since the momenta have already been calculated for you for all but one value, this problem is very simple::
\[ρ=20Ns+12Ns=2Ns+xNs=32Ns\]Solving for the missing value you'll get 30Ns.

Answer 5

bro there is no whee mentioned that the collision is elastic. .
however the x=20Ns and y=12Ns
after collision motion of x would directly propotional to the force exerted by y. . .
and motion of y would be directly proportional to the force exerted by x. . .
thus if x is directly proportional to 12Ns and its momentum after collision is 2Ns that is the system lost its 5/6th of the enrgy in the collision( 2= 12(1-5/6) 0r 12/6)
thus similarly the momentum of the y would be 1/6th of the momenta transfered by x to y that is 20Ns so the answer is 20/6Ns=10/3Ns.

Answer 6

@theyatin: You are correct...it didn't say the collision was elastic and I thought it did.

@Ashish02: Since the problem didn't specifically say it was elastic I'd go with 10/3 Ns:\[\frac{2Ns}{12Ns}=\frac{1}{6}\]\[(\frac{1}{6})20Ns=\frac{20Ns}{6}=\frac{10Ns}{3}\]