Question

1. Non-Linear Collisioin A particle A which has a mass m and travels with a speed vo along some direction i collides with a stationary particle B having mass 2m. As indicated in the figure, after the collision A moves in a direction at an angle of 30° from i and B moves in a direction at angle 50° from i . The only forces acting on the particles are those which they exert on each other. a) Apply the momentum law along a i to 30° direction perpendicular to determine the ratio of final speedsm VB /VA (1,B is the speed of B after the collision and is the speed of A after the collision) 50° 2m b) Apply the momentum law along a direction parallel to i · Use this information together with that from part (a) to find the speeds A and B after the collision. Express your answers in terms of the original speed of A, vo

Answer 1

As there is no external force on the system so the net momentum of the system will be conserved that is

momentum before collision will be equal to momentum after collission.

a)

So in direction perpendicular to i

initial momentum=0

final mometum=

mV_a'sin30-2mV_b'sin50

So

mV_a'sin30-2mV_b'sin50=0

V_b'/V_a'=sin30/2sin50

=0.326

b)

Now applying conservation of momentum in i

inital momentum= mV₀

final momentum,

mV_a'ços30+2mV_b'cos50

So

mV_a'ços30+2mV_b'cos50=mV₀.............(1)

And in part a we have found that

V_b'/V_a'=0.326....(2)

So,

simplifying equation 1

0.866V_a'+1.285V_b'=V₀

From equation 2

V_b'=0.326V_a'

V_a'=1.285V₀

V_b'=0.419V₀