Help me with this...?

Question

Help me with this...?
a ballA of mass 1 units travelling at velocity of 2 units collides with a ballB of mass 1.2 units which is at rest. take the collision as head on and perfectly elastic. now can u please find the velocities of the balls after the collision using law of cons of momentum...?

Answer 1

The answer is this --- after the elastic collision,
vB = 1.8181... units and vA = - 0.1818... units.

Here's how to show this. The relevant equations are:

1*2 + 1.2*0 = 1*vA + 1.2*vB ..(1) [Conservation of momentum], and

vB - vA = 2 ..(2)

[Elastic collision ==> relative velocity reversed. This fact allows one to avoid writing the quadratic equation that would otherwise be required from the conservation of energy.]

From (2), vB = 2 + vA ..(3)

Inserting vB from (3) into (1), 2 = vA + 1.2(2 + vA) = 2.2vA + 2.4;

so 2.2vA = - 0.4, therefore vA = - 0.4/2.2 = - 0.1818... .

Then, from (3), vB = 2 - 0.1818... = 1.8181... .

QED

A CHECK of this solution (ALWAYS advisable): (i) The relative velocity is 1.8181... - (- 0.1818...) = 2, as required for an elastic collision, and (ii) vA + 1.2vB = - 0.1818... + 2.1818... = 2, so that equation (1) is indeed satisfied. So the conditions of the problem are met by this solution.

Live long and prosper.

Answer 2

Conservation of momentum:
(1)(2) = (1)Va + (1.2)Vb
Conservation of energy:
(1)(4) = (1)Va^2 + (1.2)Vb^2
Va = 2 - 1.2Vb
1.44Vb^2 - 4.8Vb + 4 + 1.2Vb^2 = 4
2.64Vb^2 - 4.8Vb = 0
Vb(0.11Vb - .2) = 0
Vb = 1.8182
Va = 2 - (1.2)(1.8182) = - 0.1818