A 72- kg ice hockey player standing on a frictionless sheet of ice throws a 5.6- kg bowling ball horizontally with a speed of 3.9 m/s. With what speed does the hockey player recoil?
momentum before = 0
momentum after = 0
0 = 72 v + 5.6 * 3.9
v = -.303
recoils at .303 m/s
To determine the speed at which the hockey player recoils after throwing the bowling ball, we can apply the principle of conservation of momentum.
The law of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act on it. In this case, since the ice is assumed to be frictionless, we can consider the system of the hockey player and the bowling ball as closed.
The momentum of an object is the product of its mass and velocity. Mathematically, momentum (p) is given by:
p = m * v
Where:
p = momentum
m = mass of the object
v = velocity of the object
Since the bowling ball is thrown horizontally, its vertical velocity component is zero. Therefore, the horizontal momentum of the bowling ball is:
momentum of bowling ball = mass of bowling ball * velocity of bowling ball
momentum of bowling ball = 5.6 kg * 3.9 m/s
momentum of bowling ball = 21.84 kg·m/s
According to the law of conservation of momentum, the total momentum of the system before the throw is equal to the total momentum of the system after the throw. This means that the momentum of the hockey player and the bowling ball together before the throw is equal to the momentum of the hockey player after the throw.
Let's assume the velocity of the hockey player after the throw is v'. As the mass of the hockey player is given as 72 kg and initially at rest, the momentum of the hockey player before the throw is zero.
The total momentum after the throw is:
momentum after throw = mass of the hockey player * velocity of the hockey player
momentum after throw = 72 kg * v'
According to the conservation of momentum:
momentum before throw = momentum after throw
mass of bowling ball * velocity of bowling ball = mass of hockey player * velocity of hockey player
5.6 kg * 3.9 m/s = 72 kg * v'
Solving for v':
v' = (5.6 kg * 3.9 m/s) / 72 kg
v' ≈ 0.3025 m/s
Therefore, the hockey player recoils with a speed of approximately 0.3025 m/s after throwing the bowling ball.