A 15.0-gram air hockey puck sliding on a horizontal surface at a velocity of 7.00 meters per second northcollides with a 15.0-gram air hockey puck traveling at a velocity of 8.00 meters per second south. Themomentum of the system of pucks after the collision is
Question
A 15.0-gram air hockey puck sliding on a horizontal surface at a velocity of 7.00 meters per second northcollides with a 15.0-gram air hockey puck traveling at a velocity of 8.00 meters per second south. Themomentum of the system of pucks after the collision is
Solution
The momentum of a system is conserved if no external forces are acting on it. This means that the total momentum before the collision is equal to the total momentum after the collision.
Step 1: Calculate the momentum of each puck before the collision.
Momentum is calculated by multiplying the mass of an object by its velocity.
For the first puck: Momentum = mass x velocity = 15.0g x 7.00 m/s = 105.0 g.m/s north
For the second puck: Momentum = mass x velocity = 15.0g x 8.00 m/s = 120.0 g.m/s south
Step 2: Determine the total momentum before the collision.
The pucks are moving in opposite directions, so we subtract the momentum of the second puck from the first.
Total momentum before = 105.0 g.m/s - 120.0 g.m/s = -15.0 g.m/s
The negative sign indicates that the total momentum is 15.0 g.m/s to the south.
Step 3: Determine the total momentum after the collision.
According to the law of conservation of momentum, the total momentum after the collision is the same as before the collision.
So, the momentum of the system of pucks after the collision is 15.0 g.m/s to the south.
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