Question

A 110 gram hockey puck slides across the frictionless surface of an ice rink at 2.2 meters per second. It collides with a 420 gram water bottle at rest on the ice. The water bottle slides off at 0.84 meters per second in the direction of the puck's original velocity. What's the puck's final velocity?

Answer 1

The final velocity of the hockey puck is 1.44 m/s.

Step-by-step explanation:

To find the final velocity of the hockey puck, we can use the principle of conservation of momentum. Momentum is defined as the product of an object's mass and velocity. Before the collision, the total momentum of the system is equal to the momentum of the hockey puck, since the water bottle is at rest. After the collision, the total momentum is equal to the momentum of the water bottle, since the hockey puck is now at rest.

Using the equation for conservation of momentum:

m1v1i + m2v2i = m1v1f + m2v2f

Where:

• m1 is the mass of the hockey puck (110 grams or 0.11 kg)
• v1i is the initial velocity of the hockey puck (2.2 m/s)
• m2 is the mass of the water bottle (420 grams or 0.42 kg)
• v2i is the initial velocity of the water bottle (0 m/s)
• v1f is the final velocity of the hockey puck (unknown)
• v2f is the final velocity of the water bottle (0.84 m/s)

Substituting the given values into the equation and solving for v1f, we find that the final velocity of the hockey puck is 1.44 m/s.