Momentum problem due this Tuesday

  • Thread starter FizzixIzFun
  • Start date
  • Tags
    Momentum
In summary, The problem involves a 365 kg mass being released from rest on a 60 degree frictionless track at a height of 4.2 m above a horizontal surface. It collides elastically with a 480 kg mass initially at rest on the surface and the 480 kg mass slides up a similar track. Using conservation of energy, the velocity of the 480 kg mass immediately after the collision can be found by setting the gravitational potential energy of the first block equal to the kinetic energy of the second block. To find the maximum height the 480 kg mass will slide, conservation of energy can be used again by setting the kinetic energy of the second block equal to its gravitational potential energy at its highest point.
  • #1
FizzixIzFun
14
0
Momentum homework problem due this Tuesday

Homework Statement


Here's the question:
The surfaces are frictionless. The tracks are 60 degrees from horizontal. A 365 kg mass is released from rest on a track at a height 4.2 m above a horizontal surface at the foot of the slope. It collides elastically with a 480 kg mass initially at rest on the horizontal surface. The 480 kg mass slides up a similar track. The acceleration of gravity is 9.8 m/s^2. (A.) What is the speed of the 480 kg block immediately after the collision? Answer in units of m/s. (B.) To what maximum height above the horizontal surface will the 480 kg mass slide? Answer in units of m.


Homework Equations


(I'm assuming by relevant equations, you mean the ones that I think are relevant. If you mean equations given in the problem, then disregard these two equations)
GPE= mgh
KE=(1/2)mv^2


The Attempt at a Solution


(A.) I found the gravitational potential energy of the first block mgh=365 * 9.8 * 4.2 and then set that equal to (1/2)(365)(v^2) and solved for v. Is that the correct way to solve for the velocity of the 480 kg block since they collided elastically? (v=9.073 m/s)

(B.) I would assume that since the collision is elastic, the height the 480 kg mass would reach would be 4.2 (the height that the 365 kg mass started at). I feel like that is way to easy and that I must not be seeing something. (h=4.2m)
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
For (A): by using energy conservation and solving for v, you get the velocity of the 365 kg mass when it reaches the horizontal plane. Now you know the velocity of that mass before the collision. Use conservation of ________ and _______ ______ to solve for the velocities after the collision.

For (B): use energy conservation again.
 
  • #3
FizzixIzFun said:
Here's the question:
The surfaces are frictionless. The tracks are 60 degrees from horizontal. A 365 kg mass is released from rest on a track at a height 4.2 m above a horizontal surface at the foot of the slope. It collides elastically with a 480 kg mass initially at rest on the horizontal surface. The 480 kg mass slides up a similar track. The acceleration of gravity is 9.8 m/s^2. (A.) What is the speed of the 480 kg block immediately after the collision? Answer in units of m/s. (B.) To what maximum height above the horizontal surface will the 480 kg mass slide? Answer in units of m.
Hi FizzixIzFun, welcome to the forums,

Just for future reference there are Homework forums dedicated to homework questions.
FizzixIzFun said:
Here's what I figured:
(A.) I found the gravitational potential energy of the first block mgh=365 * 9.8 * 4.2 and then set that equal to (1/2)(365)(v^2) and solved for v. Is that the correct way to solve for the velocity of the 480 kg block since they collided elastically?
Your on the right track, but not quite there yet. You calculated the potential energy correctly, and you are correct to assume that this energy is transferred to the kinetic energy of 480kg object. However, you have note solved correctly for the velocity. What you have found is the velocity of the 365kg object not the 480kg object as you are required to do.
FizzixIzFun said:
(B.) I would assume that since the collision is elastic, the height the 480 kg mass would reach would be 4.2 (the height that the 365 kg mass started at). I feel like that is way to easy and that I must not be seeing something.
Indeed you are missing something. Compare the gravitational potential energy of the 360kg object with that of the 480kg.
 
  • #4
Ok, I think I got it. (A.) I use (365)gh = (1/2)(480)v^2 and solve for v. (B.) I use (1/2)(480)v^2= (480)gh and solve for h. Is that correct?
 
  • #5
Ok, I think I got it. (A.) I use (365)gh = (1/2)(480)v^2 and solve for v. (B.) I use (1/2)(480)v^2= (480)gh and solve for h. Is that correct?
 
  • #6
FizzixIzFun said:
Ok, I think I got it. (A.) I use (365)gh = (1/2)(480)v^2 and solve for v. (B.) I use (1/2)(480)v^2= (480)gh and solve for h. Is that correct?

The equation you wrote implies that the velocity of the 365 kg object equals zero after the elastic collision, which is not true.
 
  • #7
OK, I'm lost. I don't know what to do.
 
  • #8
FizzixIzFun said:
OK, I'm lost. I don't know what to do.

Ok, let's slow down. Let the 365 kg mass be m1 and the other one be m2. What we can do first is use conservation of energy to find the velocity of the mass m1 when it reaches the horizontal plane. The other mass, m2, is at rest, right? The they collide. Since the collision is elastic, you know that momentum and kinetic energy must be conserved. You must use these facts to calculate the velocity of the mass m2 after collision. (v1 has some other velocity, which isn't of interest to us.) After you do so, you can use energy conservation again to calculate the height the mass m2 climbs up to.
 
  • #9
Here's what I've come up with. Using m1gh=(1/2)m1v1^2, I solve for the velocity of m1 right before the collision. Then, I do m1v1=m2v2 and solve for v2 which would be the velocity of the block immediately after the collision. I then do (1/2)m2v2^2=m2gh and solve for h which gives me the height to which m2 rises. Please tell me I've got it this time.
 
  • #10
FizzixIzFun said:
Here's what I've come up with. Using m1gh=(1/2)m1v1^2, I solve for the velocity of m1 right before the collision. Then, I do m1v1=m2v2 and solve for v2 which would be the velocity of the block immediately after the collision. I then do (1/2)m2v2^2=m2gh and solve for h which gives me the height to which m2 rises. Please tell me I've got it this time.

Everything is OK except the equation marked red. Conservation of momentum tells you that m1v1 = m1v1' + m2v2'. Further on, you have to use conservation of kinetic energy, i.e. 1/2 m1v1^2 = 1/2m1v1'^2 + 1/2m2v2'^2. Solve the system of equations for v2' (v1' is of no interest to us.), and proceed as you already did. To see why that is so, here's a useful link: http://hyperphysics.phy-astr.gsu.edu/hbase/elacol2.html#c1" (look at the 'Head-on Elastic Collisions' part.).
 
Last edited by a moderator:
  • #11
Thank you so much for the help. I've got it now.
 

1. What is a momentum problem?

A momentum problem is a type of physics problem that involves calculating the momentum of an object or system of objects. Momentum is a measure of the amount of motion an object has, and it is calculated by multiplying an object's mass by its velocity.

2. How do I solve a momentum problem?

To solve a momentum problem, you will need to use the equation p = mv, where p is momentum, m is mass, and v is velocity. You may also need to use the principle of conservation of momentum, which states that the total momentum of a closed system remains constant.

3. What are some common real-world examples of momentum problems?

Some common examples of momentum problems include collisions between objects, explosions, and rocket launches. These scenarios involve the transfer of momentum between objects or systems.

4. What units are used to measure momentum?

Momentum is typically measured in kilogram-meters per second (kg*m/s) in the metric system. In the imperial system, it is measured in slug-feet per second (slug*ft/s).

5. How can I check if my answer to a momentum problem is correct?

You can check your answer by making sure it has the correct units and by using the principle of conservation of momentum. If your answer does not match the expected result, you may have made a mistake in your calculation.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
365
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top