Presentation on theme: "Unit 2 – Section 4 –Lesson 1 Momentum and Impulse."— Presentation transcript:
Unit 2 – Section 4 –Lesson 1 Momentum and Impulse
At this point you may have realized that momentum is related to Newton’s 2 nd Law. Remember, then And, then This is impulse, a vector quantity defined as the product of the force acting on a body and the time interval during which the force is exerted.
The Pillow and the Brick - A Question of Impulse A.A 15 N Pillow, 1m above the man B. A 15N Brick, 1m above man Both have the same mass and weight. Ignoring friction, both fall at same velocity. Therefore, both have the same momentum. Which is “better” to be hit with?
Impulse Impulse is defined as the product of the unbalanced or net force and the time that the force is acting Impulse = J = F Net x t From the pillow and brick problem, both objects have the same weight and are dropped from the same height, to hit you on top of the head (both fall at the same rate). Since the pillow is soft and flexible it strikes your head and begins to “settle” around the shape of your head before coming to a stop. The pillow takes much more time to (say 0.5s ) before it comes to a stop. The brick is very hard and does not “give” as it hits your head, it comes to a stop very quickly (perhaps in 0.01 s)
The stop time is very different in each case. How does the stop time relate to the force of the object’s impact on your head? 15 N Feather Pillow 15 N Brick Time to stop: 0.50 s Time to stop 0.010 s How much force does each have at impact? Pillow: Brick: Since each object has the same mass and velocity they have the same momentum. If the momentums are equal we consider the impulse of each collision to be equal. Suppose the impulse in each case is 60.0 N∙s Compare the impact force of each Remember: Impulse = F x ∆t
Impulse in Sports Sweet Spot explained http://news.cnet.com/8301 -13772_3-9914179-52.html A short interaction time or “time to stop” = larger applied force A longer interaction time or “time to stop” = smaller applied force
By Increasing the time of impact, the force of the impact can be decreased Practical Applications i.Modern cars are safer than older cars. Older cars have a very stiff rigid construction. During an accident it would stop quickly and large force is transferred to the car occupants. A newer car is built with a crumple zone at the from of the car. On impact the head of the car buckles and lengthens the time for the car to stop. This reduces the force on the car and the occupants.
ii. Rubber buoys are placed between a boat and the wharf as a it docks. The impulse or change of momentum at impact will be the same with or without the bumpers. By placing the buoys we increase the length of time for the boat to stop. This means that the impact force is lessened. Answer each of the following (on the Journal Sheet Provided). 1.Why should you bend your knees as you jump from the stage in a gymnasium and land on the gym floor ? 2.Why do stunt men use giant airbags when they land high-fall students ? 3.Automobile passenger air bags increase safety during accidents. Why ? 4.Which would be better for driving a nail in a board – A steel hammer or a rubber mallet? Explain in terms of Impulse Momentum.
The impulse of a collision is equal to the object’s change in momentum. Since the mass of the object nearly always remains the same as it stops only its velocity changes. This leads to the following formula: The Impulse-Momentum Theorem This part is impulse The units are N∙s 1 N∙s = (1 kg∙m/s 2 )(s) = 1 kg∙ m/s This part is change in momentum The units are kg∙m/s These units are the same as those for change in momentum.
Ex 1. A net force of 4.5 N is applied to a stationary dynamic cart with a mass of 1.15 kg. What will be velocity of the cart after 2.5 s ? F∆t = m∆v (4.5)(2.5) = (1.15)(v 2 – 0 ) 11.25 = 1.15 v 2 V 2 = 9.8 m/s 2
Ex. 2 A 0.10 kg snowball travelling at 20.0 m/s [Left] hits you in the head and stops in 0.16 s. What force does the snowball exert when it hits you? Ex. 3 A 100.0 g golf ball leaves the tee at 100.0 m/s. If the club exerts a force of 500.0 N on the ball. Determine how long the club was in contact with the ball. Ex. 4 A 180 g baseball is thrown at 30.0 m/s toward a batter whose strikes it so that the ball travels at 45 m/s in the opposite direction/ If the bat and ball are in contact for 2.0 x 10 -3 s, what is the force of the bat on the ball?
Ex. 2 A 0.10 kg snowball travelling at 20.0 m/s [Left] hits you in the head and stops in 0.16 s. What force does the snowball exert when it hits you? F∆t = m∆v (F)(0.16) = (0.10)(0 – (-20) ) 0.16 F = 2 F = 2 / 0.16 F = 12.5 = 1.3 x 10 1 N [Right] – head on snowball or - 1.3 x 10 1 N [Left] – snowball on head
Ex. 3 A 100.0 g golf ball leaves the tee at 100.0 m/s. If the club exerts a force of 500.0 N on the ball. Determine how long the club was in contact with the ball. F∆t = m∆v (500.0)(t) = (0.100)(100 – 0 ) 500 t = 10 t = 10 / 500 t = 0.02 s
Ex. 4 A 180 g baseball is thrown at 30.0 m/s toward a batter whose strikes it so that the ball travels at 45 m/s in the opposite direction/ If the bat and ball are in contact for 2.0 x 10 -3 s, what is the force of the bat on the ball? F∆t = m∆v F (2.0 x 10 -3 ) = (0.180)(-45 – 30.0 ) F (2.0 x 10 -3 ) = -13.5 F = -13.5 / 2.0 x 10 -3 F = 6750 = 6800 N away from the batter