Perfectly elastic collision?

In summary, the conversation discusses the problem of a package of mass m colliding with a previous package of mass 2m at the bottom of a frictionless chute. Assuming a perfectly elastic collision, the goal is to determine the height to which the package of mass m will rebound after the collision. The correct approach involves setting up equations for conservation of momentum and conservation of energy, either by rearranging the equations or by using the velocity of the center of mass. The final velocity for the mass m is found to be -2.6 m/s, resulting in a rebound height of 33 cm.
  • #1
mattpd1
13
0

Homework Statement


A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

http://session.masteringphysics.com/problemAsset/1073693/3/10.P42.jpg




Homework Equations






The Attempt at a Solution



So far I know that the velocity of mass m at the moment of impact is 2.6 m/s. I think I should be using the conservation of energy/momentum to solve for the final velocity of mass m. I then think I can use this velocity to solve for K, and use this to solve for the height from U=mgh. I just can't figure out how to solve the final velocities.

[tex]2.6m=mv_{1}_{f}+2mv_{2}_{f}[/tex]

[tex]\frac{1}{2}m(2.6)^2=\frac{1}{2}mv_{1}_{f}^2+mv_{2}_{f}^2[/tex]

Are these set up right? If so, how do I continue? As you can see my algebra skills may be lacking.
 
Physics news on Phys.org
  • #2
First of all I see that "m" appears everywhere. Can you simplify it ?
 
  • #3
Your result of 2.6 m/s is wrong. You can calculate this with (final kinetic energy) = (initial potential energy), so (1/2)mv^2 = mgh

Apart from this your equations are correct. For a start you can divide them by m.

There are 2 ways of doing this. The first is to rearrange the first equation to get v_1 as a function of v_2, and then substitute this in the second equation.

A better way is to find the velocity of the center of mass, and compute all velocities in the reference frame of the center of mass. The equations become

[tex] 0 = mv_{1}_{f}+2mv_{2}_{f} [/tex]

(0 because it's the center of mass frame)

[tex] KE = \frac{1}{2}mv_{1}_{f}^2+mv_{2} _{f}^2 [/tex]

where KE is the initial kinetic energy in this frame

It's then easy to see that if (v1,v2) is a solution to these equations, so is (-v1,-v2)

Finally you have to transform the velocities back to the rest frame.
 
  • #4
willem2 said:
Your result of 2.6 m/s is wrong. You can calculate this with (final kinetic energy) = (initial potential energy), so (1/2)mv^2 = mgh

Apart from this your equations are correct. For a start you can divide them by m.

There are 2 ways of doing this. The first is to rearrange the first equation to get v_1 as a function of v_2, and then substitute this in the second equation.

A better way is to find the velocity of the center of mass, and compute all velocities in the reference frame of the center of mass. The equations become

[tex] 0 = mv_{1}_{f}+2mv_{2}_{f} [/tex]

(0 because it's the center of mass frame)

[tex] KE = \frac{1}{2}mv_{1}_{f}^2+mv_{2} _{f}^2 [/tex]

where KE is the initial kinetic energy in this frame

It's then easy to see that if (v1,v2) is a solution to these equations, so is (-v1,-v2)

Finally you have to transform the velocities back to the rest frame.

Oh crap. 2.6 is the speed of the objects after an inelastic collisions. I have been using the wrong number the whole time. Let me try it the right way... Is the velocity 7.67 m/s?

GOT IT! 33cm is the rebound height.

When I did finally get the correct FINAL velocity for the mass, it was -2.6. Is this just a coincidence that this is the same as the 2.6 that I thought was the initial velocity? Anyway, thank you!
 
Last edited:
  • #5



I would first clarify the concept of a perfectly elastic collision. A perfectly elastic collision is one in which the total kinetic energy of the system is conserved before and after the collision. This means that the objects involved do not experience any permanent deformation or energy loss during the collision.

In this scenario, the package of mass m will rebound to a height equal to its initial height (3.0m) because the collision between the two packages is perfectly elastic. This is because the total kinetic energy of the system (composed of the two packages) is conserved.

To solve for the final velocities of the two packages, you can use the conservation of momentum and conservation of energy equations. Since the collision is perfectly elastic, the total kinetic energy before and after the collision will be the same. This means you can equate the two equations:

\frac{1}{2}mv_{1}_{i}^2=\frac{1}{2}mv_{1}_{f}^2+\frac{1}{2}(2m)v_{2}_{f}^2

You can then use the conservation of momentum equation to solve for one of the final velocities:

mv_{1}_{i}=(m+2m)v_{1}_{f}

Solving for v_{1}_{f} gives:

v_{1}_{f}=\frac{m}{3m}v_{1}_{i}=\frac{1}{3}v_{1}_{i}

Substituting this into the energy equation gives:

\frac{1}{2}mv_{1}_{i}^2=\frac{1}{2}m(\frac{1}{3}v_{1}_{i})^2+\frac{1}{2}(2m)v_{2}_{f}^2

Solving for v_{2}_{f} gives:

v_{2}_{f}=\frac{2}{3}v_{1}_{i}

Now that you have the final velocities, you can use the conservation of energy equation to solve for the height to which the package of mass m will rebound:

\frac{1}{2}mv_{1}_{f}^2=mgh

Solving for h gives:

h=\frac{v_{1}_{f}^2}{2g}=\frac{(\frac{1}{3}v_{1}_{i})^2}{2g}=\frac{1}{9}\frac{v_{1
 

1. What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which there is no loss of kinetic energy. This occurs when two objects collide and bounce off each other without any deformation or energy being converted into other forms, such as heat or sound.

2. How is momentum conserved in a perfectly elastic collision?

In a perfectly elastic collision, the total momentum of the system (the two objects colliding) is conserved. This means that the total momentum before the collision is equal to the total momentum after the collision.

3. Can real-life collisions be perfectly elastic?

No, in real-life collisions, some amount of kinetic energy is always lost due to factors such as friction, deformation, and sound. Therefore, perfectly elastic collisions are an idealized concept used in physics to simplify calculations and understand the basic principles of collisions.

4. How is the coefficient of restitution related to perfectly elastic collisions?

The coefficient of restitution (e) is a measure of how much kinetic energy is conserved in a collision. In a perfectly elastic collision, the value of e is 1, indicating that all kinetic energy is conserved. In real-life collisions, the value of e is always less than 1, as some energy is lost.

5. Can a perfectly elastic collision occur between objects of different masses?

Yes, a perfectly elastic collision can occur between objects of different masses. In this case, the lighter object will experience a larger change in velocity compared to the heavier one, but the overall total momentum and kinetic energy of the system will still be conserved.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
717
  • Introductory Physics Homework Help
Replies
6
Views
868
  • Introductory Physics Homework Help
Replies
4
Views
992
  • Introductory Physics Homework Help
Replies
20
Views
843
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Introductory Physics Homework Help
Replies
15
Views
216
  • Introductory Physics Homework Help
Replies
21
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
772
Back
Top