Conservation of Momentum of two trolleys

AI Thread Summary
Two trolleys are set to collide, with Trolley 1 moving at 4 m/s and Trolley 2 at 1 m/s, having masses of 2 kg and 4 kg, respectively. The total momentum before the collision is calculated as 12 kg·m/s, confirming momentum conservation. After the collision, the combined mass is 6 kg, leading to a post-collision velocity of 2 m/s. The kinetic energy after the collision is then calculated using the formula, yielding 12J as the correct answer. The discussion highlights an alternative method to find kinetic energy using momentum directly.
Charles W
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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
 
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Edit: I should read all the post better ...
Charles W said:
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.
 
Charles W said:
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.
 
Doc Al said:
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?
 
Charles W said:
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.
 
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Orodruin said:
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!
 
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