Momentum Is Mass Times Velocity: P = M V

Momentum and Impulse For better understanding we’ll do another example: h = 2 m m = 30 g = 0.03 kg  Momentum is mass times velocity: p = m v vector! unit: (p) = kg m/s

Let’s go back to Newton’s second law: F = ma. Actually, Newton formulated his second law as:  Force = time rate of change of momentum

Δp is the change in momentum produced by the force F in time Δt. If the mass doesn’t change, then

both eggs fall the same distance, so the velocity F = ma is actually a special case of Newton’s second law and of both eggs just before impact is: it

can not be applied to the situations in which mass can change. Impact: before impact u = 6 m/s Another very useful form of Newton’s 2. law: just after impact is: v = 0

 Impulse F∆t will produce change in momentum Δp ∆p = mv – mu = – 0.18 kg m/s

F∆t = ∆p Δp = mv - mu In both cases momentum is reduced to zero during impact/interaction with the floor. But the time of interaction F∆t is called the impulse of the force F. is different. In the case of concrete, time is small while in the action of a force F over time Δt case of pillow, the stopping time is greatly increased. If you look at the impulse -momentum relation F∆ t = ∆p, you units: (F∆t) = Ns Ns = kg m/s see that for the same change in momentum (– 0.18 kg m/s in this case), if the time is smaller the ground must have exerted REMEMBER: Although we write F for simplicity, we actually greater force on the egg. And vice versa. The pillow will exert mean Fnet , because only Fnet and not individual forces can smaller force over greater period of time. change momentum (by producing an acceleration)  Getting smart and smarter by knowing physics:

Achieving the same change in momentum over Often you want to reduce the momentum of an object to zero a long time requires smaller force and over a but with minimal impact force (or injury). How to do it? Try to short time requires greater force. maximize the time of interaction; this way you’ll decrease the stopping force. • Car crash on a highway, where there’s either a concrete wall or a barbed-wire fence to crash into. Which to choose? Naturally, the wire fence – your momentum will be decreased by the same amount, so the impulse to stop you is the same, but with the wire fence, you extend the time of impact, so decrease the force. • Bend your knees when you jump down from high! Try keeping your knees stiff while landing – it hurts! (only try for a small jump, otherwise you could get injured…) Bending the knees extends the time for momentum to go to zero, by about 10- 20 times, so forces are 10-20 times less. • Safety net used by acrobats, increases impact time, decreases the forces. • Catching a ball – let your hand move backward with the ball after contact… • bungee jumping • Wearing the gloves when boxing versus boxing with bare fists. 2 particles is conserved - remains constant, provided Sometimes you want to increase the force over a short time there is no resultant external force. • This is how in karate, Such a system is called an “isolated system”. an expert can break a stack mathematically: of bricks with a blow of a hand: p is momentum of the system p = p + p Bring in arm with tremendous 1 2

speed (large momentum), that p = p m v + m v = m u + m u is quickly reduced on impact after before 1 1 2 2 1 1 2 2 with the bricks. The shorter the time, the larger Certain situations (collisions, explosions, ejections) do not allow the force on the bricks. detailed knowledge of forces. The problem is that it is difficult to see exactly how to apply Newton’s laws . One cannot easily  Impulse due to a time varying force measure neither forces involved in the collision nor acceleration (velocity appears to be instantaneously acquired). Formulas we had are for constant force. The law of conservation of momentum gives us an easy and What if the force changes over time ∆t? elegant way to predict the outcome without knowing forces involved in process. It is much easier to measure velocities and The graph shows the variation with time of the masses before and after strong interaction. force on the football of mass 0.5 kg. WE CAN APPLY THE LAW OF CONSERVATION OF MOMENTUM TO COLLISIONS AND EXPLOSIONS (EJECTIONS) IF DURING INTERACTION THE NET EXTERNAL FORCE IS EITHER ZERO OR CAN BE NEGLECTED.

Example: baseball is struck with a bat – duration of the collision is about 0.01 s, and the average force the bat exerts on the ball is several thousand Newtons what is much greater than the force of gravity, so you can ignore it.

 Examples how to use law of conservation of Change in momentum, Δp, in time Δt, is the area momentum in the case of ejection or explosion. under the graph force vs. time. • A 60.0-kg astronaut is on a space walk when her tether line F∆t = ∆p → ∆p = 5 kg m/s breaks. She is able to throw her 10.0-kg oxygen tank away from ∆p = m∆v → ∆v = 10 m/s the shuttle with a speed of 12.0 m/s to propel herself back to v = u + ∆v the shuttle. What is her velocity?

Collisions – conservation of momentum

• collisions can be very complicated + • two objects bang into each other and exert strong

forces which are very hard to measure over short time intervals • fortunately, we can predict the future without going pafter = pbefore into pesky details of force. 0 = m1 v1 + m2 v2 • What will help us is the law of conservation of momentum: 0 = 60.0 v1 + 10.0 (12.0) = 0  Law of Conservation of Momentum v1 = − 2.0 m/s moving toward shuttle particle 1 and particle 2 collide with one another. • Very similar case is spaceship propulsion which is actually velocities just before velocities just after example of conservation of momentum. The momentum collision, before they interaction (collision) gained by fuel ejected in the backward direction must be touch each other balanced by forward momentum gained by the spaceship.

The total momentum of a system of interacting 3

(6) (1 ) + (2) (—2 ) = (6 + 2 ) v 6 — 4 = 8 v v = 0.25 m/s in the direction of the large fish before lunch The negative momentum of the small fish is very effective in slowing the large fish. c. Small fish swims toward the large fish at 3 m/s. • Similar examples are: recoil of the firing gun, recoil of the firing cannon, ice-skater’s recoil, throwing of the package from the boat etc.

p before lunch = p after lunch (6) (1 ) + (2) (—3 ) = (6 + 2 ) v 6 — 6 = 8 v v = 0

Two stationary ice skaters push off fish have equal and opposite momenta. Zero momentum before Both skaters exert equal forces on each other however, lunch would equal zero momentum after lunch, and both fish the smaller skater acquires a larger speed (due to would come to a halt. larger acceleration) than the larger skater. Momentum is conserved! d. Small fish swims toward the large fish at 4 m/s. +

pafter = pbefore

0 = p1 + p2

p1 = – p2 (6) (1 ) + (2) (—4 ) = (6 + 2 ) v  Example how to use law of conservation of 6 — 8 = 8 v v = — 0.25 m/s momentum in the case of collisions. The minus sign tells us that after lunch the two-fish system moves There are two fish in the sea. A 6 kg fish and a 2 kg fish. in a direction opposite to the large fish’s direction before lunch. The big fish swallows the small one. What is its velocity immediately after lunch

• A red ball traveling with a speed of 2 m/s along the x-axis hits the a. the big fish swims at 1 m/s toward and swallows eight ball. After the collision, the red ball travels with a speed of o the small fish that is at rest. 1.6 m/s in a direction 37 below the positive x-axis. The two balls have equal mass. At what angle will the eight ball fall in the side pocket? What is the speed of the blue (8th) ball after collision.

p before lunch = p after lunch

Mu1 + mu2 = (M+m)v (6 kg)(1 m/s) + (2 kg)(0 m/s) = (6kg + 2 kg)v 6 kg.m/s = (8 kg) v v = 0.75 m/s

in the direction of the large fish before lunch 0 in x – direction m u1 + 0 = m v1 cos 37 + m v2 cos q2 0 v2 cos q2 = u1 - v1 cos 37 = 0.72 m/s b. Suppose the small fish is not at rest but is swimming toward (1) the large fish at 2 m/s. 0 in y – direction 0 = - m v1 sin 37 + m v2 sin θ2 0 v2 sin θ2 = v1 sin 37 = 0.96 m/s (2)

0 direction of v2 ; (2)/(1) tan θ2 = 1.33 θ2 = 53 p before lunch = p after lunch 4 0 (2) → v2 = 0.96 / sin 53 v2 = 1.2 m/s