Momentum and Conservation of Momentum

[PhysicPsychic]

Whew, I finished the kinetic and power problems thanks to the help of marlon and arildno, but now I have some questions on momentum. Hopefully this will help others who need help with the concept.


Firstly, I have some questions on basic principle to get down(I'm hoping I got these right as to not mess up later with false knowledge, so please correct me):

The impulse experienced by a body is equivalent to the body's change in _____?

I had no idea what the question meant so I said force.

A large moving ball collides with a stationary small ball. The momentum ______.

I put down "the momentum of the large ball decreases, and the momentum of the small ball increases."

Two objects with different masses collide and bounce back after the collision. Before the collision, the two objects were moving at velocities equal in magnitude but opposite in direction. After the collision,________.

Well, since this is an elastic collision and both total momentum and total kinetic energy reamain constant, I said "After the collision, they both had the same momentum."

Two skaters stand facing each other. One skater's mass is 60 kg, and the other's mass is 72 kg. If the skaters push away from each other,_______.

I always thought that since they were pushing away from each other, mass doesn't really matter, so I said "their momentum is equal but opposite."

In a two body collision,________.

Not knowing what type of collison, I just said that "both momentum and kinetic energy are conserved."

Whew, that's enough of the basics(which I hope I got right), and onto the word problems!


A bullet with a mass of 5.00 x 10³ is loaded into a gun. The loaded gun has a mass of .52 kg. The bullet is fired, causing the empty gun to recoil at a velocity of 2.1 m/s. What is the velocity of the bullet?

Well, since the total momentum of the problem is constant, and the intial quantities are 0, I'm guessing the final momentum is going to equal zero.

Also, since the gun is recoiling, the gun must be going in the negative direction if I make the bullet's direction positive.
so:

m₁*v₁,f + m₂*v₂,f = (.005 kg *v₁,f) + (.52 kg * -2.1 m/s)

which would equal

.005 kg (v₁,f) - 1.1 kg*m/s = 0

add 1.1...

solve for v₁,f by dividing 1.1 kg*m/s by .005 kg.

kg cancels and I am left with the bullet having a velocity of 220 m/s.

An object with a mass of .10 kg makes an elastic head-on collision with a stationary object with a mass of .15 kg. The final velocity of the .10 kg object after the collision is -.045 m/s. What was the initial velocity of the .10 kg object?

Hmmm, an elastic collision. I will probably use the same conservation of momentum equation, but I am missing the final velocity of the second object.

How can I find that second velocity?

 

[HallsofIvy]

Whew, I finished the kinetic and power problems thanks to the help of marlon and arildno, but now I have some questions on momentum. Hopefully this will help others who need help with the concept.


Firstly, I have some questions on basic principle to get down(I'm hoping I got these right as to not mess up later with false knowledge, so please correct me):

The impulse experienced by a body is equivalent to the body's change in _____?

I had no idea what the question meant so I said force.

Why would force change? The force of gravity, for example, doesn't change. Impulse is, by definition, the change in momentum.

A large moving ball collides with a stationary small ball. The momentum ______.

I put down "the momentum of the large ball decreases, and the momentum of the small ball increases."

Assuming you are only talking about magnitude of momentum, not direction (momentum is a vector), yes, the small ball was stationary and so by moving at all, its momentum increases. In order to conserve momentum, that of the large ball must decrease.

Two objects with different masses collide and bounce back after the collision. Before the collision, the two objects were moving at velocities equal in magnitude but opposite in direction. After the collision,________.

Well, since this is an elastic collision and both total momentum and total kinetic energy reamain constant, I said "After the collision, they both had the same momentum."

Did the problem say the collision was elastic? And what do you mean "they both had the same momentum"? The two objects will have the same momentum after the collision or they will each have the same momentum as they had before the collision? The only thing you can really say here is that the total momentum remains the same.

Two skaters stand facing each other. One skater's mass is 60 kg, and the other's mass is 72 kg. If the skaters push away from each other,_______.

I always thought that since they were pushing away from each other, mass doesn't really matter, so I said "their momentum is equal but opposite."

Why wouldn't mass matter? "pushing away from each other" just means that the force on both sides is equal but F= ma so with different masses, they will have different accelerations. In any case, the total momentum initially is 0 so after you will have 60v₁- 72v₂= 0. That's not enough to solve for the speeds of course because you don't know what the force was.

In a two body collision,________.

Not knowing what type of collison, I just said that "both momentum and kinetic energy are conserved."

Not knowing what type of collision, you assume it is elastic? All you can say is that momentum is conserved.

Whew, that's enough of the basics(which I hope I got right), and onto the word problems!


A bullet with a mass of 5.00 x 10³ is loaded into a gun. The loaded gun has a mass of .52 kg. The bullet is fired, causing the empty gun to recoil at a velocity of 2.1 m/s. What is the velocity of the bullet?

Well, since the total momentum of the problem is constant, and the intial quantities are 0, I'm guessing the final momentum is going to equal zero.

You don't need to guess! Of course the initial momentum is 0, nothing was moving. (By the way, I'll bet that the mass of the bullet was 5.00x 10^-3!)

Also, since the gun is recoiling, the gun must be going in the negative direction if I make the bullet's direction positive.
so:

m₁*v₁,f + m₂*v₂,f = (.005 kg *v₁,f) + (.52 kg * -2.1 m/s)

which would equal

.005 kg (v₁,f) - 1.1 kg*m/s = 0

add 1.1...

solve for v₁,f by dividing 1.1 kg*m/s by .005 kg.

kg cancels and I am left with the bullet having a velocity of 220 m/s.

I didn't check the arithmetic but the method is right.

An object with a mass of .10 kg makes an elastic head-on collision with a stationary object with a mass of .15 kg. The final velocity of the .10 kg object after the collision is -.045 m/s. What was the initial velocity of the .10 kg object?

Hmmm, an elastic collision. I will probably use the same conservation of momentum equation, but I am missing the final velocity of the second object.

How can I find that second velocity?

Since this is an elastic collision you know two things: both momentum and kinetic energy are conserved.
Letting v_1i and v_1f be the initial and final speeds of the .10 kg object, and v_2i and v_2f the initial and final speeds of the 0.15 kg object, you know that v_2i = 0 (it was "stationary") and v_1f= -.045. That leaves v_1i and v_2f unknown. Set up the momentum and conservation equations and solve those two equations for the two unknowns.

 

[pmrazavi]

I'm really curious to know what you got from the exam.. :)