Momentum

__Tuesday__




**__Wednesday:__**
Textbook pages 208-214 (all sets of problems)

__**Thursday:**__
Notes on conservation of momentum and types of collisions. (see above powerpoint). We also played with the marble cars and tracks to figure out conceptually how conservation of momentum applies to cars of differing masses and velocities.

__ANSWER KEY FOR BOOKWORK (I'm only including work for reasonably difficult problems)__

page 209:

1) 2482 kgm/s 2a) 121 kgm/s 2b) 94.5 kgm/s 2c) 26.6 kgm/s 3) 46.3 m/s

page 211

1) 375 N 2) We need to find the velocity that he hits the water with first, because that will be his momentum when he hits the water. So, using our motion (or conservation of E) equations we know that an object falling from a distance of 3 meters will be 7.67 m/s. So, momentum is 629 kgm/s. That drops to zero so the impulse is -629 kgm/s. So, divide that by time to get the force, which is -1144 N. It has to be negative because it's a stopping force. 3) Impulse = Pf - Pi. So, impulse = -16 kgm/s. (the final velocity was -22 m/s and the initial was 18 m/s) 4a) 9 m/s (Ft = change in momentum and the initial momenturm was zero) 4b) initial momentum = 4.5 kgm/s. The impulse provided by the force is 12 kgm/s in the opposite direction. So, the final momentum is -7.5 kgm/s. So, the final velocity is -15 m/s.

page 214

1a) doubled 1b) quadrupled 2a) 31.0 m/s (thanks Gracie!) 2b) the bulllet is 3375 J and the baseball 69.67 J, so the bullet has more. (Thanks Gracie!) 3) Not if the large force is given over a much smaller time comparitively. But, all things equal, a large force WOULD cause a greater change in momenturm because the change is equal to Ft 4) Momentum is a measure of the object's "mass in motion", or how much force it would take to stop the object. Impulse is the change in momentum due to some force either acting with the motion or opposing it. 5a) 2.52 kgm/s 5b) 126 N

__Monday__



__Wednesday__





__Thursday__





__**PRACTICE TEST (Conceptual Questions)**__
1) What are the units of momentum and impulse? 2) How is impulse related to momentum? 3) If you provide an impulse to an object, in what circumstance will it slow down? In what circumstance will it speed up? In what circumstance will it turn? (think about it) 4) Explain in your own words the Law of Conservation of Momentum 5) Two children are standing still on ice and push off each other. Each has a momentum after the push. Doesn't this violate the conservation law? It was zero before the push. 6) Describe how the conservation law works with elastic and inelastic collisions. 7) When objects collide inelastically in 2-D, why can't you simply make a triangle with total Vx and Vys in order to find out the final velocity of the new object? Why doesn't that work? 8) In general, for elastic collisions with stationary objects, when will the first object bounce back? In general, when will it follow the stationary object after the collision? 9) During the lab it became apparent that momentum was NOT conserved in the collisions. Why was that? 10) How are momentum and kinetic energy similar? How are they different?

__**PROBLEMS**__

1) A 5 kg car moving at 20 m/s collides with a 10 kg car traveling in the same direction at 3 m/s. Calculate the resulting velocities for BOTH cars for both elastic and inelastic scenarios. 2) A 4 gram bullet is fired at 340 m/s into a stationary 2 kg block of wood. Assuming the floor and the wood block have a coefficient of kinetic friction equal to 0.45, how far will the block slide before stopping? What is the work done by the force of friction? How many seconds will the block slide? 3) Instead of a flat surface, assume the block of wood is at the bottom of a 20 degree ramp with the same coefficient of friction. How high up (vertically) will the block go? What will the distance traveled by the block before stopping? What will the time be? (draw it out. You already know how to calculate all this stuff from our unit on force and energy) 4) A object with mass m is shot horizontally from a cliff with height h with initial velocity v. There is a cart with mass 5m waiting to catch the object at the bottom of the cliff (placed at the right horizontal distance). What is the equation for the velocity of the cart after it catches the object?



3) to solve for time use Pf - Pi = Ft where F is the combined force of kinetic friction and gravity (mgsin). Time = 0.091 seconds 4) This is somewhat of a trick question, because it's far easier than you might think. Since the horizontal velocity of a falling object never changes, when the object lands in the cart it has horizontal velocity V. Therefore, the Vf of the cart + object is: Vf = (mv)/(6m), which reduces to Vf = 1/6(v)