# Conservation of momentum problems

I was gone when my teacher assigned this, and now its spring break and i have no notes. Im sure they arent hard and all i am missing are equations. If anyone could help me that would be great :D

1. a .04 kg bullet is fired at a velocity of 1200 m/s from a 6kg gun. what is the recoild velocity of the gun

I think ive figured this out
V = -mv / M
so v = -(.04)(1200)/(6)
would be 8 m/s in the opposite direction :)

2. a 40 kg girl sits on a 15kg sled holding two 10kg bowling balls. Assuming the sled starts at rest on a frictionless horizontal surface, she throws the first ball backwards with a velocity of 4 m/s. then the second ball with the same velocity. What is the sled speed after the first ball is thrown? after the second ball?

Initial mass = 75 kg
after first ball 65 kg
after 2nd ball 55 kg

so initially you take v = -mv / M
v = -(10)(4)/(65) = .62 m/s

and for the 2nd ball you would use
v = -(10)(4)/(55) = .72 m/s

What i dont understand is how you calculate the final velocity for this problem.

3. A rocket can expel 200kg of mass per second at a velocity of 4000 m/s if the launce of the rocket requires an acceleration of 2.0 m/s, what is the greatest mass the rocket may be at the time of launch? (assume net mass of rocket is constant)

if i work this like a gun problem i get
V = mv / M
2 = (200)(4000)/ M
2M = (200)(4000)
M = 400000kg

4. a 8000 kg train travels north along a track at 3.0 m/s and collides with a stationary car of mass 13000 kg. if the cars couple and stick together, what is the final velocity of the two car system?

this one i think i know, using the equation
m1v1 + m2v2 = (m1 + m2) Vf
(8000)(3) + (13000)(0) = (8000 + 13000) Vf
24000 + 0 = 21000 Vf
Vf = 24000 / 21000 = 1.14 m/s
just want to make sure :)

5. a neutron (mass = 1.67 x 10tothe-27 kg( collides with a "resting" helium atom (mass = 6.64 x 10tothe-27 kg) at a celoxity of 2 x 10tothe6 m/s. The final velocity of the neutron is 1.8 x 10tothe6 m/s. What is the velocity of the helium atom, assuming the collision is in one dimension?

Thanks for any help i get :)

Last edited:

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dynamicsolo
Homework Helper
1. a .04 kg bullet is fired at a velocity of 1200 m/s from a 6kg gun. what is the recoild velocity of the gun

would it be 8 m/s? just a guess.
Why guess? Write out the conservation of momentum equation and you'll see why your answer works.

I also say this because the forum rules are that you show your work on a problem and then someone helps you with it.

2. a 40 kg girl sits on a 15kg sled holding two 10kg bowling balls. Assuming the sled starts at rest on a frictionless horizontal surface, she throws the first ball backwards with a velocity of 4 m/s. then the second ball with the same velocity. What is the sled speed after the first ball is thrown? after the second ball?
This is similar to problem 1, but there are two stages.

3. A rocket can expel 200kg of mass per second at a velocity of 4000 m/s if the launce of the rocket requires an acceleration of 2.0 m/s, what is the greatest mass the rocket may be at the time of launch? (assume net mass of rocket is constant)
What does the rate of change of momentum with respect to time represent? (Hint: what would the units be?) This would give you an idea of how to start this one.

4. a 8000 kg train travels north along a track at 3.0 m/s and collides with a stationary car of mass 13000 kg. if the cars couple and stick together, what is the final velocity of the two car system?

this one i think i know, using the equation
m1v1 + m2v2 = (m1 + m2) Vf
(8000)(3) + (13000)(0) = (8000 + 13000) Vf
24000 + 0 = 21000 Vf
Vf = 24000 / 21000 = 1.14 m/s
just want to make sure :)
Correct.

5. a neutron (mass = 1.67 x 10tothe-27 kg) collides with a "resting" helium atom (mass = 6.64 x 10tothe-27 kg) at a velocity of 2 x 10tothe6 m/s. The final velocity of the neutron is 1.8 x 10tothe6 m/s. What is the velocity of the helium atom, assuming the collision is in one dimension?
If you were able to do problem 4, you should be able to set up this one.

Thanks for the reply, its actually pretty hard to find equations on the net, so i posted here.

Ive figured out problem 1 and edited my post, and im still having problems with #2, im trying to work through it. Do you just take an average of the numbers?

Again thanks.

dynamicsolo
Homework Helper
2. a 40 kg girl sits on a 15kg sled holding two 10kg bowling balls. Assuming the sled starts at rest on a frictionless horizontal surface, she throws the first ball backwards with a velocity of 4 m/s. then the second ball with the same velocity. What is the sled speed after the first ball is thrown? after the second ball?

Initial mass = 75 kg
after first ball 65 kg
after 2nd ball 55 kg

so initially you take v = -mv / M
v = -(10)(4)/(65) = .62 m/s

Since the girl on the sled holding the bowling balls is initially at rest, the initial (linear) momentum is zero. The momentum after the first ball is thrown is given by

(40+15+10) · V + 10 · 4 = 0 , since the total momentum must still be zero.

Solving this for V gives the result you found (though it should be
-0.615 m/sec if we call the direction in which the bowling ball is thrown, positive).

For the second part, the momentum of the girl on the sled with the remaining ball is not zero, since she will continue to recoil on the frictionless surface. But we know that the momentum of the first thrown ball is 10 · 4 = +40 kg-m/sec , so the momentum of girl et al. is now -40 kg-m/sec (since the total must be zero.) So you set up a second conservation of momentum equation for the new situation, with the second ball thrown at 4 m/sec relative to the sled (this is presumably what the problem means by the second ball being thrown "with the same velocity" -- actually, this part of the sentence is poorly expressed).

3. A rocket can expel 200kg of mass per second at a velocity of 4000 m/s if the launce of the rocket requires an acceleration of 2.0 m/s, what is the greatest mass the rocket may be at the time of launch? (assume net mass of rocket is constant)

if i work this like a gun problem i get
V = mv / M
2 = (200)(4000)/ M
2M = (200)(4000)
M = 400000kg
Watch your units on this one! You are given a rate of mass expulsion out of the rocket body times the velocity (relative to the rocket body). The product is

(200 kg/sec)·(4000 m/sec) , which is a "rate of momentum transfer" from the rocket.

What units is this in, that is to say, what other physical quantity has those units? How might that be relevant to what the question asks for?

Don't forget to do Problem 5 also...