Question

Object A of mass M is initially at rest on a flat, smooth frictionless surface. Object B, which has twice the mass of A, is traveling with speed V before it collides elastically with A. Immediately after the collision, both objects move off at angles (theta)>0 with respect to the original direction of B. Calculate the value of the angle.

Object A of mass M is initially at rest on a flat, smooth frictionless surface. Object B, which has twice the mass of A, is traveling with speed V before it collides elastically with A. Immediately after the collision, both objects move off at angles (theta)>0 with respect to the original direction of B. Calculate the value of the angle. [Hint: Note that the collision is elastic.] .

[Hint: Note that the collision is elastic.] .

Answer 1

initial momentum = (2*M * V) i
( as only B has a velocity of 'V')

finally ,let the velocity of ball A and B   be vA and vB respectively

momentum of ball A =M* vA( cos theta i - sin theta j)

momentum of ball B = 2M* vB( cos theta i + sin theta j )

so,
conserving the momentum,
M* vA( cos theta i - sin theta j) +2M* vB( cos theta i + sin theta j ) = (2*M * V) i

so,( cancelling j components)
M*vA = 2MvB ,
so,
vA= 2 vB .......................(1)

now,( equating i components)

so,
vA cos theta + 2 vB cos theta = 2 V
also,
using (1)

vA( 2*cos theta ) = 2V
so,
vA cos theta = V .................(2)

for elastic collision, energy must be conserved ,

so ,
1/2*2M * V^2 = 1/2 * M* vA^2 + 1/2 * 2M* vB^2

from (1)
2* V^2 = vA^2 + vA^2/2
so,
V^2 = 3vA^2 /4

using (2)
vA^2 * (cos theta )^2 = vA^2*( 3/4)
so,
(cos theta )^2 =( 3/4)
so, cos theta = sqrt(3)/2

so,
theta = 30 degrees
so the angle is of 30 degrees