Slide 1 / 133 Slide 2 / 133 AP Physics 1 Momentum 2015-12-02 www.njctl.org Slide 3 / 133 Slide 4 / 133 Table of Contents Click on the topic to go to that section Momentum · · Impulse-Momentum Equation · The Momentum of a System of Objects Momentum · Conservation of Momentum · Types of Collisions · Collisions in Two Dimensions Return to Table of Contents Slide 5 / 133 Slide 6 / 133 Momentum Defined Momentum Defined Newton’s First Law tells us that objects remain in motion with a constant velocity unless acted upon by an external force (Law of Define a new quantity, momentum ( p ), that takes these Inertia). observations into account: momentum = mass × velocity In our experience: When objects of different mass travel with the same velocity, the p =mv · one with more mass is harder to stop. When objects of equal mass travel with different velocities, the · faster one is harder to stop. click here for a introductory video on momentum from Bill Nye!
Slide 7 / 133 Slide 8 / 133 SI Unit for Momentum Momentum is a Vector Quantity There is no specially named unit for momentum - so there is an Since: opportunity for it to be named after a renowned physicist! mass is a scalar quantity · velocity is a vector quantity · We use the product of the units of mass and velocity. the product of a scalar and a vector is a vector · and: momentum = mass × velocity · mass x velocity therefore: Momentum is a vector quantity - it has magnitude and direction kg ⋅ m/s Slide 9 / 133 Slide 9 (Answer) / 133 1 Which has more momentum? 1 Which has more momentum? A A large truck moving at 30 m/s A A large truck moving at 30 m/s B A small car moving at 30 m/s B A small car moving at 30 m/s A Answer Both have the same speed, but C Both have the same momentum. C Both have the same momentum. the truck has the larger mass [This object is a pull tab] Slide 10 / 133 Slide 10 (Answer) / 133 2 What is the momentum of a 20 kg object moving to the 2 What is the momentum of a 20 kg object moving to the right with a velocity of 5 m/s? right with a velocity of 5 m/s? m= 20 kg Answer v = 5 m/s p = mv = (20 kg)(5 m/s) = 100 kg ⋅ m/s [This object is a pull tab]
Slide 11 / 133 Slide 11 (Answer) / 133 3 What is the momentum of a 20 kg object with a velocity 3 What is the momentum of a 20 kg object with a velocity of 5.0 m/s to the left? of 5.0 m/s to the left? m= 20 kg Answer v = −5 m/s p = mv = (20 kg)(−5 m/s) = −100 kg- m/s [This object is a pull tab] Slide 12 / 133 Slide 12 (Answer) / 133 4 What is the velocity of a 5 kg object whose momentum is 4 What is the velocity of a 5 kg object whose momentum is −15 kg-m/s? −15 kg-m/s? m= 5 kg p = −15 m/s Answer p = mv v = p/m v= (-15 kg-m/s)/5 kg v = -3 m/s [This object is a pull tab] Slide 13 / 133 Slide 13 (Answer) / 133 5 What is the mass of an object that has a momentum of 5 What is the mass of an object that has a momentum of 35 kg-m/s with a velocity of 7 m/s? 35 kg-m/s with a velocity of 7 m/s? v= 7 m/s p = 35 m/s Answer p = mv m = p/v m= (35 kg-m/s)/(7 m/s) m = 5 kg [This object is a pull tab]
Slide 14 / 133 Slide 15 / 133 Change in Momentum Suppose that there is an event that changes an object's momentum. from p 0 - the initial momentum (just before the event) · by Δp - the change in momentum · Impulse-Momentum Equation to p f - the final momentum (just after the event) · The equation for momentum change is: Return to Table of Contents Slide 16 / 133 Slide 17 / 133 Momentum Change = Impulse Momentum change equation: Newton's First Law tells us that the velocity (and so the momentum) of an object won't change unless the object is affected by an external force. Look at the above equation. Can you relate Newton's First Law to the Δp term? Δp would represent the external force. Slide 18 / 133 Slide 19 / 133 SI Unit for Impulse There no special unit for impulse. We use the product of the units of force and time. force x time N ⋅ s Recall that N=kg ⋅ m/s 2 , so N ⋅ s=kg ⋅ m/s 2 x s This is also the unit for momentum, which is a good = kg ⋅ m/s thing since Impulse is the change in momentum.
Slide 20 / 133 Slide 20 (Answer) / 133 6 There is a battery powered wheeled cart moving towards 6 There is a battery powered wheeled cart moving towards you at a constant velocity. You want to apply a force to you at a constant velocity. You want to apply a force to the cart to move it in the opposite direction. Which the cart to move it in the opposite direction. Which combination of the following variables will result in the combination of the following variables will result in the greatest change of momentum for the cart? Select two greatest change of momentum for the cart? Select two answers. answers. Answer A Increase the applied force. A Increase the applied force. A, B B Increase the time that the force is applied. B Increase the time that the force is applied. C Maintain the same applied force. C Maintain the same applied force. D Decrease the time that the force is applied. D Decrease the time that the force is applied. [This object is a pull tab] Slide 21 / 133 Slide 21 (Answer) / 133 7 From which law or principle is the Impulse-Momentum 7 From which law or principle is the Impulse-Momentum equation derived from? equation derived from? A Conservation of Energy. A Conservation of Energy. B Newton's First Law. B Newton's First Law. Answer C Newton's Second Law. C Newton's Second Law. C D Conservation of Momentum. D Conservation of Momentum. [This object is a pull tab] Slide 22 / 133 Slide 22 (Answer) / 133 8 Can the impulse applied to an object be negative? Why 8 Can the impulse applied to an object be negative? Why or why not? Give an example to explain your answer. or why not? Give an example to explain your answer. Students type their answers here Students type their answers here Yes, since Impulse is a vector quantity, it has magnitude and direction. Direction can be described Answer as negative. An example would be if a baseball is thrown in the positive direction, and you are catching it with a mitt. The mitt changes the baseball's momentum from positive to zero - hence applying a negative impulse to it. [This object is a pull tab]
Slide 23 / 133 Slide 24 / 133 Effect of Collision Time on Force Every Day Applications Impulse = F (∆t) = F (∆t) Impulse = F (∆t) = F (∆t) The inverse relationship of Force and time interval leads to many interesting applications of the Impulse-Momentum Since force is inversely equation to everyday experiences such as: proportional to Δt, changing the F (newtons) ∆t of a given impulse by a small car structural safety design · amount can greatly change the car air bags · force exerted on an object! landing after parachuting · martial arts · hitting a baseball · catching a basebal · ∆t (seconds) Slide 25 / 133 Slide 26 / 133 Every Day Applications Car Air Bags I=FΔt=Δp I=FΔt=Δp Let's analyze two specific cases from the previous list: In the Dynamics unit of this course, it was shown how during an accident, seat belts protect passengers from the effect of car air bags · Newton's First Law by stopping the passenger with the car, and hitting a baseball · preventing them from striking the dashboard and window. Whenever you have an equation like the one above, you have to decide which values will be fixed and which will be varied to They also provide another benefit explained by the Impulse- determine the impact on the third value. Momentum equation. But, this benefit is greatly enhanced by the presence of air bags. Can you see what this benefit is? For the car air bags, we'll fix Δp, vary Δt and see its impact on F. For the bat hitting a ball, we'll fix F, vary Δt and see the impact on Δp. Slide 27 / 133 Slide 28 / 133 Car Air Bags Car Air Bags I=FΔt=Δp I=FΔt=Δp The seat belt also increases the time interval that it takes the Δp is fixed, because as long as the passenger remains in the passenger to slow down - there is some play in the seat belt, car, the car (and the passengers) started with a certain velocity that allows you to move forward a bit before it stops you. and finished with a final velocity of zero, independent of seat belts or air bags. The Air bag will increase that time interval much more than the seat belt by rapidly expanding, letting the passenger strike it, Rearranging the equation, we have: then deflating. Earlier it was stated that for the Air bag example, Δp would be fixed and Δt would be varied. So, we've just increased Δt. Why is Δp fixed? F represents the Force delivered to the passenger due to the accident.
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