Solving an Elastic Collision Problem with Given Force-Time Data

In summary, the collision between a 10-kg mass moving at an initial velocity V1 of 6 m/s and a stationary 2-kg mass results in a final velocity of 4 m/s for the 10-kg mass and a final velocity of 6 m/s for the 2-kg mass. The net force acting on each object during the collision can be represented by a force-time graph, with a peak force of 10000N at 3 milliseconds. The impulse during the collision is calculated to be 20 N*s using the momentum equation I = F/deltaT. The conservation of momentum and energy equations can be used to find the initial and final velocities of the objects.
  • #1
the7joker7
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Homework Statement



A 10-kg mass moving at a speed of V[tex]_{1}[/tex] in a positive direction collides elastically with a 2-kg mass which is initially at rest. The magnitude of the net force that acts on each object during the collision is shown in the figure. Find the initial velocity, V[tex]_{1}[/tex], of the 10-kg mass prior to collision, the final velocity, v[tex]_{3}[/tex], of the 10-kg mass, and the final velocity, v[tex]_{4}[/tex], of the 2-kg mass.

I can't post the figure up, but it's a force with respect to time chart. It starts with nothing, then starts to spike upwards at a constant rate from time 1 millisecond to time 3 milliseconds, where it peaks at 10000N, then goes back down at the same rate, hitting 0N at 5 milliseconds. I've already figured the Impulse is 20 N*s from this.

Homework Equations



p = mv

I = F[tex]/delta[/tex]T

[tex]/delta[/tex]p = mv[tex]_{final}[/tex] - mv[tex]_{initial}[/tex]

mv[tex]_{before collision}[/tex] = the some of mv[tex]_{after collision}[/tex]

Possibly assorted others.

The Attempt at a Solution



So far, I know the Impulse is 20 NS.

Impulse = Momentum, I do believe.

So

20 = (10)V

Making Velocity of the first object initially = 2 m/s, correct?

That would answer the first question, but I'm not totally sure where to go from there.
 
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  • #2
Here's a quick idea of what I tried to do...lemme know if it looks right.

20 = (10)V[tex]_{3}[/tex] + 2(V[tex]_{4}[/tex])
20 = .5(10)V[tex]_{3}[/tex][tex]^{2}[/tex] + .5(2)V[tex]_{4}[/tex][tex]^{2}[/tex]

Those are the conservation of momentum and energy equations.

I solved the momentum one for V[tex]_{3}[/tex] and got .55279 m/s. I can post steps if anyone wants to see.

Then I plugged that into the energy one and got 1.7061 m/s. Sound about right?

It looks like it MIGHT be accurate, but the sums of mv before and the sums of mv after aren't equal.

I got 2(10) = 20 for before and 10(.55279) + 2(1.7061) = 8.94 for after. =/
 
Last edited:
  • #3
my answers:

v1 + V3 = 10
2V1 = 12
V1 = 6 m/s
V3 = 4 m/s
 
  • #4
How is V1 6m/s?

Am I doing the impulse equation wrong?
 

Related to Solving an Elastic Collision Problem with Given Force-Time Data

1. What is an elastic collision?

An elastic collision is a type of collision in which both kinetic energy and momentum are conserved. This means that before and after the collision, the total kinetic energy and momentum of the system remain the same.

2. What is the difference between an elastic collision and an inelastic collision?

In an elastic collision, there is no loss of kinetic energy, while in an inelastic collision, some kinetic energy is lost in the form of heat, sound, or deformation of the objects involved.

3. How do you calculate the velocities of the objects after an elastic collision?

The velocities of the objects after an elastic collision can be calculated using the conservation of momentum and kinetic energy equations. These equations take into account the masses and initial velocities of the objects.

4. Can an elastic collision occur between two objects of different masses?

Yes, an elastic collision can occur between two objects of different masses. This is because the conservation of momentum and kinetic energy equations take into account the masses of the objects.

5. What is an example of an elastic collision in real life?

An example of an elastic collision in real life is the collision between two billiard balls on a pool table. In this case, the total kinetic energy and momentum of the system are conserved, and the balls bounce off each other with no loss of energy.

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