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Impulse A change in momentum How to change momentum • Any object that has momentum is going to be hard to stop • To stop an object a force needs to be applied to the object, against the object’s motion and for a certain period of time • It takes a greater force or amount of time to stop an object with a large momentum • The momentum of the object is changed since the velocity is changed Changes in momentum in real life • In a football game, the defensive players stop the momentum of the offensive players so the offense does not score • When driving, we apply the brakes which is a force to stop the car and change the velocity of the car Impulse • Impulse = change in momentum • The formula for impulse is: • Impulse = Δp = F * t • Δp = impulse in N*s (change in momentum) • F = force in Newtons (N) t = time in seconds (s) • reminder p = mv so Δp = m *Δv Collision • In a collision, the object experiences an impulse • The impulse is equal to the change in momentum • The object will speed up or slow down or change direction due to the force for a given amount of time Elastic Collisions • When a ball rebounds off a wall there will be a change in velocity and a change in momentum, that is the impulse • The greater the rebound effect the greater the acceleration, momentum change and impulse • A rebound has direction and speed change Elastic Collisions • If a rebound has the same speed (and momentum) before and after the collision, it is called an elastic collision • Elastic collisions usually have a large velocity change, a large momentum change and a large force Practice – fill in the blanks p = mv Δp = F * Δt Δp = m * Δv Force F Time t Impulse Momentum Δp Change Δp Mass m Velocity Change Δv N s N*s kg m/s 10 -4 1 0.010 2 0.100 3 0.010 4 -20000 5 -200 kg*m/s -40 10 -200 50 -200 1.0 -8 50 Sample Problem • A .50kg cart is pulled with a 1N force for 1 second and a .50kg cart is pulled with a 2N force for .50seconds. Which cart has the greatest acceleration and impulse? Cart 1 a = F/m so a = 1N/.50kg = 2 m/s2 impulse = F * t = 1N * 1s = 1N*s cart 2 a = F/m so a = 2N/.5kg = 4m/s2 impulse = F * t = 2N * .5s = 1N*s • Cart 2 has a greater acceleration but both carts have the same change in momentum or impulse Sample Problem 2 • A hockey player applies an average force of 80N to a .25kg hockey puck for a time of .10 seconds, what is the impulse of the hockey puck? • Δp = F * t • Δp = 80N x .10s = 8N*s