A 65.0 kg ice-skater standing on frictionless ice throws a 0.15 kg snowball horizontally at a speed of 32.0 m/s. at what speed d
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Answer:
The speed of 250 g ball after collision is 14 m/s.
Explanation:
mass of first ball, m = 550 g = 0.55 kg
initial velocity of first ball, u = 8 m/s
mass of second ball, m' = 250 g = 0.25 kg
initial velocity of second ball, u' = - 8 m/s
Let the speed of the balls is v and v' after the collision.
Use conservation of momentum
m u + m' u' = m v + m' v'
0.55 x 8 - 0.25 x 8 = 0.55 v + 0.25 v'
0.55 v + 0.25 v' = 2.4 ..... (1)
Use the formula of coefficient of restitution,
For elastic collision, e = 1
By solving (1) and (2)
v = - 2 m/s and v' = 14 m/s