What Determines the Velocity of a Dropped Weight on a Toy Car?

[Gib Z]

Ok guys I'm having a really simple problem, its part of a school project.

I have a piece of string, on one end is a 50gram weight, the other is a toy car.

The toy car is going to be on the table, and the weight is going to be held by me until I decide to let go. I know dropping the weight will make the car move.

I know the velocity of the car is the same as the velocity of the weight (the string is going to be fully strechted the whole time).

What I want to do is to multiply the velocity of the car and therefore the weight by 2. How would I go about doing this? Do I need to include other factors like height of table or should I change the weight to double the speed?

I think $E_g =mgh$, $v^2=2as$(u=0), or something else may come useful here but I can't do it.

NOTE: Ignore friction.

Thanks guys

 

[Doc Al]

What I want to do is to multiply the velocity of the car and therefore the weight by 2.

Does the car have some initial speed? Or does it start from rest? (I'm not clear on what you want to do.)

In any case, you need to consider conservation of energy. The system (car & mass) has some initial energy (PE + KE) before you release the mass. Ignoring dissipative forces, it will have the same total energy as the mass falls (until it hits the floor). The decrease in the PE of the falling mass will equal the increase in KE of the system.

 

[Gib Z]

The car is at rest at the start, and so it the weight. When I drop the weight, the car will move. I wanted to know what to do to the weight to make the velocity of the car two times as much.

This is a question I am asking for a friend but I couldn't do it either. I just realized the question isn't great because the velocity isn't constant, its accelerating with gravity. But if anyone has any thing that sort of works please tell.Doc Al- So should i equate $mg(h_0-h)=\frac{1}{2}mv^2$?

 

[Doc Al]

The car is at rest at the start, and so it the weight. When I drop the weight, the car will move. I wanted to know what to do to the weight to make the velocity of the car two times as much.

Twice zero is still zero! Better rethink what you want.

Doc Al- So should i equate $mg(h_0-h)=\frac{1}{2}mv^2$?

Not exactly. Call the falling mass m1 and the car m2. So:
$$m_1g \Delta h = \frac{1}{2}(m_1 + m_2)v^2$$

The KE of both masses increases.

 

[Gib Z]

Sorry I realized 2*0 =0, which is why i said the question is bad as its accelerating rather than constant acceleration..sorry for wasting your time Ill tell him tomorrow the question was flawed. Thanks!