Question

A snooker ball with a mass of 0.3kg moving at 2.5m/s collides with another ball of 0.2kg mass and moving in the same direction with a speed of 1.20m/s . Assuming the collision is elastic, calculate the speed and direction of each ball after collision.

Answer 1

same momentum before and after

0.3*2.5 + 0.2*1.2 = 0.3 u + 0.2 v

but elastic so same kinetic energy too

(1/2)*.3*2.5^2+(1/2)*.2*1.2^2
=(1/2)*.3*u^2+(1/2)*.2*v^2

Answer 2

momentum is conserved

and with an elastic collision, energy is conserved

capital letters - larger ball
small letters - smaller ball

MV1 + mv1 = MV2 + mv2
... (.3 * 2.5) + (.2 * 1.2) = .3 V2 + .2v2

1/2 MV1^2 + 1/2 mv1^2 = 1/2 MV2^2 + 1/2 mv2^2
... (.15 * 6.25) + (.144) = .15 V2^2 + .1 v2^2

two equations...two unknowns...algebra time

both balls continue moving in the same direction

Answer 3

Given:
M1 = 0.3kg, V1 = 2.5 m/s.
M2 = 0.2kg, V2 = 1.2 m/s.

Momentum before = Momentum after:
M1*V1 = M2*V2 = M1*V3 + M2*V4.
0.3*2.5 + 0.2*1.2 = 0.3*V3 + 0.2*V4,
Eq1: 0.3*V3 + 0.2*V4 = 0.99.

Kinetic Energy Eq.:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2.5(0.3-0.2) + 0.4*1.2)/(0.3+0.2),
V3 = (0.25 + 0.48)/0.5 = 1.46 m/s.

In Eq1, replace V3 with 1.46 and solve for V4.