Pre AP Physics Chapters 6 and 7 Momentum and Gravity

Problems (150)

Solve the problems below on your own paper. Each problem must be solved on its own page. Be sure to use the 3-column method. Naturally, you are encouraged to use your formula/equation sheet. be sure that your boxed solutions are clearly marked with 'a', 'b', or 'c' etc.

1. A subatomic particle with a mass of 2.1 x 10^-26 kg initially at rest decays into three particles. One of the particles, with a mass of 7.0 x 10^-27 kg goes in the positive ‘x’ direction at 7.0 x 10⁶ m/s, while the other with a mass of 1.16 x 10^-26 kg goes off in the positive ‘y’ direction at 5.5 x 10⁶ m/s.

a) What is the momentum of the particle that went in the ‘x’ direction?

b) What is the momentum of the particle that went in the ‘y’ direction?

c) Draw a vector diagram of the event

d) What is the magnitude of the momentum of the third particle?

e) What is the direction of the third particle?

f) What is the mass of the third particle?

2. Sally, who has a mass of just 50 kg is attracted, at least gravitationally, to Bill, who has a substantial mass of 65 kg. Indeed the force of this mutual attraction between Sally and Bill is 6.7 x10^-8 N.

a) How far apart are Sally and Bill?

b) If Sally and Bill get three times closer to each other what will the magnitude of their mutual gravitational attraction be? You will get full credit for this if you do NOT use the 3-column method to find the answer.

c) If Sally and Bill were separated so they were five time farther away from each other what will the magnitude of their mutual gravitational attraction be? You will get full credit for this if you do NOT use the 3-column method to find the answer.

3. A steel ball bearing with a mass of 0.5 kg is thrown upwards vertically with an initial velocity of 3 m/s.

a) What is the magnitude of the momentum of the SBB just as it leaves the hand?

b) What is the magnitude and direction of the momentum of the SBB when it is halfway up to its maximum height?

c) What is the magnitude and direction of the momentum of the SBB when it is halfway down from its maximum height?

d) What is the kinetic energy of the SBB just as it leaves the hand?

e) What is the kinetic energy when the SBB is half way up to its maximum height? You will get twice the earned credit for this question if you solve it without using the 3CM. Just be sure that you state how it is that you know you are correct.

f) Without using the 3CM, determine the magnitude of the gravitational potential energy of the SBB at the top of its path? State how you know you are correct.

Pre AP Physics Chapters 6 and 7 Momentum and Gravity

Answer Section

PROBLEM

1. ANS:

a and b) Momentum is simply mv. Multiply the mass and velocity of the two particles to find their momentum.

c) Your vector diagram should look something like this:

d) To find the resultant you need only to find the hypotenuse of the vector triangle form part ‘c’. For that you will use

e) Find the mass by invoking the Law of Conservation of mass and subtracting the sum of the two particles from the mass of the original particle:

f) To find theta simply use your favorite trig function. I like to use tangent:

2. ANS:

a) To find the distance between Sally and Bill we use the Universal Gravity equation: solved for ‘d’: .

b) If Sally and Bill are three times closer to each other the Gravitational attraction between them will be 3² or 9 times greater. So, 9 x 6.7 x10^-8 N will be your answer.

c) IF Sally and Bill are five times farther away from each other Gravity will get 5² or 25 times less, so 6.7 x10^-8 N/25 will give you your answer.

3. ANS:

a) Momentum is just mv, so p=mv.

b) It is tempting to think that half way up the momentum will be half, but the SBB is accelerating! At halfway up it will not be going half speed!

To find the momentum, then, you need to find the velocity at the halfway mark.

To find the velocity at the halfway mark you need to know the distance to the halfway mark.

To know the distance to the halfway mark you need the whole vertical distance.

So, first we find the whole vertical distance using KE = GPE or , solve this for ‘h’. Using half of that value use this equation again only this time solve it for ‘v’. Now we can find the momentum.

c) Of course halfway down the magnitude of the momentum will the same as it was half way up. The direction will, of course, be opposite.

d) Simply apply the KE equation:

e) This is where this problem gets interesting! While the momentum was not half at halfway up the energy is half at halfway up! The reason this must be so is because the KE is being converted into GPE as the SBB goes up. The sum of the KE and GPE must always be the same original KE the SBB had, the KE you calculated in part ‘d”. If the GPE is half at the halfway mark going down, as it must be since it is a function of ‘h’, the KE must, therefore, also be at the halfway mark.

f) KE at the bottom must equal GPE at the top, after all energy is conserved.