A 0.105-kg hockey puck moving at 26 m/s is caught and held by a 72-kg goalie at rest. With what speed does the goalie slide on the ice?
Mpuck*Vpuck = (Mpuck+Mgoalie)*Vfinal
Vfinal = (0.105/72.1)*26 m/s
It's called momentum conservation
It assumes that goalie is not in contact with the goal
To find the speed at which the goalie slides on the ice after catching the hockey puck, we can use the principle of conservation of linear momentum.
The momentum of an object is given by the product of its mass and velocity. According to the conservation of momentum, the total momentum before and after an event remains constant if there are no external forces acting on the system.
Let's denote the mass of the hockey puck as m1 (0.105 kg), the initial velocity of the puck as v1 (26 m/s), the mass of the goalie as m2 (72 kg), and the final velocity of the goalie as v2 (unknown).
The initial momentum of the system is the momentum of the hockey puck before it was caught, given by:
Initial momentum (before catching the puck) = m1 * v1
The final momentum of the system is the momentum of the system after the puck is caught, given by:
Final momentum (after catching the puck) = (m1 + m2) * v2
Since momentum is conserved, we can equate the initial and final momentum:
m1 * v1 = (m1 + m2) * v2
Simplifying the equation, we can solve for v2:
v2 = (m1 * v1) / (m1 + m2)
Plugging in the given values:
v2 = (0.105 kg * 26 m/s) / (0.105 kg + 72 kg)
Calculating this expression, we find:
v2 ≈ 0.03 m/s
Therefore, the goalie slides with a speed of approximately 0.03 m/s after catching the hockey puck.