Momentum

What is Momentum? Momentum – tendency of objects to keep going in the same direction with the same speed Depends on mass and velocity Has direction

The momentum of a ball depends on its mass and velocity Ball B has more momentum than ball A

The momentum of a ball depends on direction too

Which has greater momentum: a train at rest or a moving skateboard? Question Which has greater momentum: a train at rest or a moving skateboard? The moving skateboard has a greater momentum because it has a velocity. The train at rest has a greater mass, but zero velocity.

Conservation of Momentum Conservation of Momentum – momentum cannot be created or destroyed, only transferred p before = p after VIDEO

Collisions Collision – occurs when two or more objects hit each other During a collision, momentum is transferred from one object to another because Momentum is conserved.

Collisions Elastic Collision – objects bounce off each other Inelastic Collision – objects stick to each other

Practice A 1200 kg car drives west at 25 m/s for 3 hours. What is the car’s momentum? (1200 kg) × (25 m/s) = 30,000 [kg m/s] west

Practice (1300 kg) × (13.5 m/s) = 17,550 kg m/s north Motorcycle A car is traveling at a velocity of 13.5 m/s north on a straight road. The mass of the car is 1,300 kg. A motorcycle passes the car at a speed of 30 m/s. The motorcycle (with rider) has a mass of 350 kg. Calculate and compare the momentum of the car and motorcycle. Car (1300 kg) × (13.5 m/s) = 17,550 kg m/s north Motorcycle (350 kg) × (30 m/s) = 10,500 kg m/s north

Practice (100 kg) × (20 m/s) = 2,000 kg m/s downhill A child is on a sled moving down a hill at 20 m/s. The combined mass of the sled and child is 100 kg. What is the momentum of the child and sled? (100 kg) × (20 m/s) = 2,000 kg m/s downhill

Practice (15 kg m/s) ÷ (30 m/s) = .5 kg A ball moving at 30 m/s has a momentum of 15 kg m/s. What is the mass of the ball? (15 kg m/s) ÷ (30 m/s) = .5 kg

Practice A boat is sitting at rest a short distance from a dock. A child in the boat throws a 6.4 [kg] package out horizontally with a speed of 10 [m/s] towards the dock. Calculate the total momentum of the system after the package is thrown. p (before) = (26 [kg] + 45 [kg] + 6.4 [kg]) * 0 [m/s] = 0 [kg*m/s] So, p (after) = 0 [kg*m/s] The total momentum after the package is thrown is the SAME as before it was thrown. WHY? The momentum is the same before and after due to the Law of Conservation of Momentum. The package definitely has a positive momentum after it is thrown, but the boat & child together will have a negative momentum after the package is thrown because they will be moving backward with a speed of -0.9 m/s. This means when you add the final momentum of the package to the final momentum of the boat/child, the TOTAL momentum will equal zero! Calculation: p before = p after (77.4 kg)*(0 m/s) = (6.4 kg)*(10 m/s) + (71 kg) * v 0 [kg*m/s] = 64 [kg*m/s] + (71 kg) * v -64 [kg*m/s] = (71 kg) * v V = (-64 [kg*m/s]) / (71 kg) V = -0.901 [m/s]