Answer:
1 m/s
Explanation:
From the law of conservation of momentum,
Total momentum before collision = Total momentum after collision
mu + m'u' = mv + m'v'...................... Equation 1
Where m = mass of the first ball m' = mass of the second ball, u = initial velocity of the first ball, u' = initial velocity of the second ball, v = final velocity of the first ball, v' = final velocity of the second ball.
Note: Let the direction of the first ball be positive.
Given: m = 0.5 kg, u = 9.0 m/s, m' = 1 kg, u' = - 11 m/s ( opposite direction to the first ball), v = - 15 m/s ( bounces backward).
Substitute into equation 1
0.5(9) + 1(-11) = 0.5(-15) + 1(v')
Solving for v'
4.5-11 = -7.5 + v'
-6.5 = -7.5 + v'
v' = 7.5-6.5
v' = 1 m/s.
Hence the speed of the second ball after collision = 1 m/s
