Chapter 7 Linear Momentum

Objectives Define impulse, and relate to momentum. Give examples of how both the size of the force and length of time applied affect the change in momentum. Solve momentum and impulse problems. Warm-Up Knowing how to find momentum, how do you think one can find the change in momentum?

Engagement / Exploration Demo – egg and sheet https://www.youtube.com/watch?v=7RSUjxiZnME

Exploration – Interpret the diagram It turns out that having a net force is not enough to cause a change in the motion of an object.  A net force must actually be present for some instant of time.  A huge force acting for zero seconds accomplishes nothing.  In fact, a small force acting for a long time can be as effective as a huge force acting for a short time.  Take the case of stopping a car.  You can abruptly stop a car by stomping on the brakes or you can stop it gradually by lightly applying the brakes.   In either case the car gets stopped.  When we multiply the average force by the time it acted for, we find that in all three cases shown above the answer is the same.   The product of average net force and change in time is called Impulse.  The formula for it actually comes from a little manipulation of Newton's Second Law.   This can be seen in the formula manipulation shown below.

Impulse F t = Δ p F t = m Δv F t = m (vf - vi) When an object experiences a net force, its momentum will change! Impulse is a change in momentum! J = Δ p Impulse = Force * Time J = F t F t = Δ p F t = m Δv F t = m (vf - vi)

Impulse changes Momentum A greater impulse exerted on an object A greater change in momentum OR Impulse = Change in momentum OR Greek symbol “Delta” Means “the change in…” Impulse = Δ(mv) Impulse can be exerted on an object to either INCREASE or DECREASE its momentum.

Impulse-Momentum Theorem The theorem states that the impulse acting on the object is equal to the change in momentum of the object Impulse=change in momentum (vector!) If the force is not constant, use the average force applied

Air Bags The air bag increases the time of the collision It will also absorb some of the energy from the body It will spread out the area of contact decreases the pressure helps prevent penetration wounds

Case 1: Increasing Momentum Examples: Hitting a golf ball: Apply the greatest force possible for the longest time possible. Accelerates the ball from 0 to high speed in a very short time. The impulse of the bat decelerates the ball and accelerates it in the opposite direction very quickly. Baseball and bat: Video: Changing Momentum – Follow Through

Case 2: Decreasing Momentum It takes an impulse to change momentum, and Remember … Impulse = F x t If you want to stop something’s motion, you can apply a LOT of force over a short time, Or, you can apply a little force over a longer time. Remember, things BREAK if you apply a lot of force to them.

Case 3: Decreasing Momentum over a Short Time If the boxer moves away from the punch, he extends the time and decreases the force while stopping the punch. If he moves toward the punch, he decreases the time and increases the force The airbag extends the time over which the impulse is exerted and decreases the force. Hitting the bricks with a sharp karate blow very quickly maximizes the force exerted on the bricks and helps to break them.

7-3 Collisions and Impulse During a collision, objects are deformed due to the large forces involved. Since , we can write The definition of impulse: (7-5)

Same Impulse If an object experiences a change in momentum, how can you minimize the force on the object? Extending the time, there by minimizing the force. J= F t J= F t

7-3 Collisions and Impulse Since the time of the collision is very short, we need not worry about the exact time dependence of the force, and can use the average force.

7-3 Collisions and Impulse The impulse tells us that we can get the same change in momentum with a large force acting for a short time, or a small force acting for a longer time. This is why you should bend your knees when you land; why airbags work; and why landing on a pillow hurts less than landing on concrete.

Impulse examples Follow through increases the time of collision and the impulse                       I small large

Problem Bobo hits a 0.050 kg golf ball, giving it a speed of 75 m/s. What impulse did he impart on the ball? Assume the initial speed of a ball was 0 m/s. J = change in momentum (mv) J = 3.75 kg m/s

Questions Pick one to run into, a brick wall or a haystack. Catch a baseball, what do you do? Jump off a table, what do you do? On which surface is a dropped glass less likely to break: carpet or sidewalk? Why do boxers use short, fast jabs?

Conservation of Momentum If no net external force (same as saying “no net impulse”) acts on a system, the system’s momentum cannot change. Momentum = 0 before the shot And after the shot Cannon’s momentum Shell’s momentum (equal and opposite)

Opposite the person’s momentum Example 7-6 Advantage of bending knees when landing! a) m =70 kg, h =3.0 m Impulse: p = ? Ft= p = m(0-v) First, find v (just before hitting): KE + PE = 0 m(v2 -0) + mg(0 - h) = 0  v = 7.7 m/s Impulse: p = mv p = -540 N s Just before he hits the ground  Just after he hits the ground   Opposite the person’s momentum

Advantage of bending knees when landing! Impulse: p = -540 N s m =70 kg, h =3.0 m, F = ? b) Stiff legged: v = 7.7 m/s to v = 0 in d = 1 cm (0.01m)! vavg = (½ )(7.7 +0) = 3.9 m/s Time t = d/v = 2.6  10-3 s F = p/t = 540 Ns/2.6  10-3 s = 2.1  105 N (Net force upward on person) From free body diagram, F = Fgrd - mg  2.1  105 N Fgrd =F + mg = 2.1  105 N + (70kg x 9.80 m/s/s)  2.1  105 N Enough to fracture leg bone!!!

Advantage of bending knees when landing! Impulse: p = -540 N s m =70 kg, h =3.0 m, F = ? c) Knees bent: v = 7.7 m/s to v = 0 in d = 50 cm (0.5m) vavg = (½ )(7.7 +0) = 3.8 m/s Time t = d/v = 0.13 s F = p/t = 4.2  103 N (Net force upward on person) From free body diagram, F = Fgrd - mg  4.9  103 N Leg bone does not break!!!

Practice Problem A 57 gram tennis ball falls on a tile floor. The ball changes velocity from -1.2 m/s to +1.2 m/s in 0.02 s. What is the average force on the ball? Identify the variables: Mass = 57 g = 0.057 kg Δvelocity = +1.2 – (-1.2) = 2.4 m/s Time = 0.02 s using FΔt= mΔv F x (0.02 s) = (0.057 kg)(2.4 m/s) F= 6.8 N

p1 = mv1 = -22500 kg m/s, p2 = mv2 = 3900 kg m/s Example: Crash Test Crash test: Car, m = 1500 kg, hits wall. 1 dimensional collision. +x is to the right. Before crash, v = -15 m/s. After crash, v = 2.6 m/s. Collision lasts Δt = 0.15 s. Find: Impulse car receives & average force on car. Assume: Force exerted by wall is large compared to other forces Gravity & normal forces are perpendicular & don’t effect the horizontal momentum  Use impulse approximation p1 = mv1 = -22500 kg m/s, p2 = mv2 = 3900 kg m/s J = Δp = p2 – p1 = 2.64  104 kg m/s (∑F)avg = (Δp/Δt) = 1.76  105 N

Closure - Car Crash Would you rather be in a head on collision with an identical car, traveling at the same speed as you, or a brick wall? Assume in both situations you come to a complete stop. Take a guess http://techdigestuk.typepad.com/photos/uncategorized/car_crash.JPG

Car Crash (cont.) Everyone should vote now Raise one finger if you think it is better to hit another car, two if it’s better to hit a wall and three if it doesn’t matter. And the answer is…..

Car Crash (cont.) The answer is… It Does Not Matter! Look at FΔt= mΔv In both situations, Δt, m, and Δv are the same! The time it takes you to stop depends on your car, m is the mass of your car, and Δv depends on how fast you were initially traveling.

Homework Chapter 7 problems 15, 16, 17, 19, 20 Kahoot 7-3