Conservation of Momentum of two trolleys

[Charles W]

Homework Statement

There are two trolleys traveling in the same direction which are about to collide.

Trolley 1 is traveling at 4 metres per second and has a mass of 2kg
The other, Trolley 2, (which is in front) is moving a 1 metre per second and has a mass of 4kg.

After the impact, they move off together

What is the total kinetic energy of the trolleys after the collision:

1. 1.3J
2. 12J
3. 18J
4. 19J

Homework Equations

Linear Momentum (kgms^-1) = Mass (kg) * Velocity (ms^-1)

The Attempt at a Solution

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I could be approaching it completely the wrong way, but I have tried to work out the momentum of each trolley prior to the collision:

Momentum of Trolley 1 = 2kg * 4 metres per second = 8kgms^-1
Momentum of Trolley 2 = 4kg * 1 metre per second = 4kgms^-1

As I understand it, as momentum is conserved, the combined momentum once they have collided is 12kgms^-1

Any help would be much appreciated

 

[Orodruin]

Edit: I should read all the post better ...

Momentum of Trolley 1 = 2kg * 4 metres per second = 8kgms^-1
Momentum of Trolley 2 = 4kg * 1 metre per second = 4kgms^-1

As I understand it, as momentum is conserved, the combined momentum once they have collided is 12kgms^-1

Correct.

 

[Doc Al]

I could be approaching it completely the wrong way, but I have tried to work out the momentum of each trolley prior to the collision:

Momentum of Trolley 1 = 2kg * 4 metres per second = 8kgms^-1
Momentum of Trolley 2 = 4kg * 1 metre per second = 4kgms^-1

As I understand it, as momentum is conserved, the combined momentum once they have collided is 12kgms^-1

So far, so good. Now use that to find the speed after the collision.

 

[Charles W]

So far, so good. Now use that to find the speed after the collision.

Thanks! So using the equation I used earlier:

Linear Momentum (kgms^-1) = Mass (kg) * Velocity (ms^-1)

Rearranging this equation, Velocity = Momentum/Mass

Therefore, by my reckoning, velocity = 12/6 = 2 metres per second

Using this, can I then use Kinetic Energy = 1/2mv^2, to give a Kinetic Energy of 12J (option 2)? Is that correct?

 

[Orodruin]

Using this, can I then use Kinetic Energy = 1/2mv^2, to give a Kinetic Energy of 12J (option 2)? Is that correct?

Yes, it is correct. I would like to point out that there is also no need for you to compute the velocity. You could equally well note that, since $p = mv$, the kinetic energy can also be expressed as $E_k = p^2/(2m)$ (this can also be useful to remember) and you only need to use your computed momentum and the total mass.

 

[Charles W]

Yes, it is correct. I would like to point out that there is also no need for you to compute the velocity. You could equally well note that, since $p = mv$, the kinetic energy can also be expressed as $E_k = p^2/(2m)$ (this can also be useful to remember) and you only need to use your computed momentum and the total mass.

Thank you very much Orodruin! I haven't seen that equation before - will definitely come in handy!