A 80- kg ice hockey player standing on a frictionless sheet of ice throws a 6.2- kg bowling ball horizontally with a speed of 2.1 m/s. With what speed does the hockey player recoil?
momentum is conserved.
m1*v1+m2V2=0
thanks
To find the speed at which the ice hockey player recoils, we can use the principle of conservation of momentum. According to this principle, the total momentum before and after the interaction must be equal.
The momentum of an object is given by its mass multiplied by its velocity. Therefore, the momentum of the hockey player and the bowling ball before the throw is:
Initial momentum = (mass of hockey player) × (velocity of hockey player) + (mass of bowling ball) × (velocity of bowling ball)
Now, we know that the hockey player is initially at rest (velocity = 0), so the initial momentum simplifies to:
Initial momentum = 0 + (mass of bowling ball) × (velocity of bowling ball)
Given:
- Mass of hockey player (m1) = 80 kg
- Mass of bowling ball (m2) = 6.2 kg
- Velocity of bowling ball (v2) = 2.1 m/s
Using the principle of conservation of momentum, the total momentum before the throw should be equal to the total momentum after the throw. Let's assume the velocity at which the hockey player recoils is v1:
Initial momentum = Final momentum
(0) + (m2 × v2) = (m1 × v1) + (0)
Since the hockey player was initially at rest, the initial momentum term becomes zero.
Now, we can solve for v1, the velocity of the hockey player after the throw:
m2 × v2 = m1 × v1
(6.2 kg) × (2.1 m/s) = (80 kg) × v1
v1 = (6.2 kg × 2.1 m/s) / (80 kg)
v1 ≈ 0.16275 m/s
Therefore, the ice hockey player recoils with a speed of approximately 0.16275 m/s.