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.
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.