A 0.3kg ball travelling at 0.5m/s, due North strikes a stationary 0.9kg ball and rebounds at 0.20m/s due South. Calculate velocity of the second ball if the collision is elastic.

Answer 1

First, we need to use the principle of conservation of momentum to solve this problem. The equation for conservation of momentum is:

m1u1 + m2u2 = m1v1 + m2v2

Where:
m1 = mass of the first ball = 0.3kg
u1 = initial velocity of the first ball = 0.5m/s
m2 = mass of the second ball = 0.9kg
u2 = initial velocity of the second ball (which is 0 as it is stationary)
v1 = final velocity of the first ball = 0.2m/s
v2 = final velocity of the second ball (to be calculated)

Plugging the given values into the equation, we get:

(0.3kg)(0.5m/s) + (0.9kg)(0) = (0.3kg)(0.2m/s) + (0.9kg)(v2)
0.15 + 0 = 0.06 + 0.9v2

Rearranging the equation, we get:

0.9v2 = 0.15
v2 = 0.15 / 0.9
v2 = 0.1667 m/s

Therefore, the velocity of the second ball after the collision is 0.1667 m/s.