# MCAT question on energy

## Homework Statement

OBjects A and B are placed on a vertical spring. A has twice the mass as B. If the spring is depressed and released, propelling the objects into the air, then A will:

a) rise 1/4 as high as b
b) rise 1/2 as high as b
c) rise the same amount as b
d) rise twice as much as b

## Homework Equations

1/2kx², 1/2mv², mgh, projectile equations, etc.

## The Attempt at a Solution

I really dont understand the logic in the solution (answer is C). It would make sense to me that the energy from the spring is transfered to kinetic, and then to gravitational potential energy for the masses. In this case, the mass of the objects would make a difference in determining the height (since its "mgh".)

The only reason I can think is that there is more energy transfered from the spring to the heavier block.

If it were just 1 block on the spring, its simply 1/2kx² = mgh, and you plug and chug. In this sense, if each object were sprung up seperately, the lighter one would definitly go higher.

Basically my problem deals with the logic in having both objects on the same spring, and how that affects the calculations/result of energy transfer.

Thanks

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Doc Al
Mentor

They would reach the same height.

Id like to understand why though, in terms of energy transfer.

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you could think about it as an elastic explosion

Dick
Homework Helper
They would reach the same height.

Id like to understand why though, in terms of energy transfer.
In terms of energy transfer the initial velocity of each mass is the same. Call it v. Then the initial kinetic energy (1/2)*m*v^2 is equal to the final potential energy m*g*h. If you solve for h you should notice that the mass cancels.

In terms of energy transfer the initial velocity of each mass is the same. Call it v. Then the initial kinetic energy (1/2)*m*v^2 is equal to the final potential energy m*g*h. If you solve for h you should notice that the mass cancels.
why is V the same? because they can be treated as one body?

Dick
Homework Helper
It's because the objects will fly off of whatever platform is accelerating them upward as soon as the acceleration of the platform becomes greater than g in a downward direction. So the criterion for flying off is independent of mass.

"It's because the objects will fly off of whatever platform is accelerating them upward as soon as the acceleration of the platform becomes greater than g in a downward direction. So the criterion for flying off is independent of mass. "

I am having a similar problem as the OP. How do you prove that the objects leave the spring surface at the same velocity?

The analogy with throwing two objects into the air only works if one assumes that the objects leave the person's hands at the same speed (which would mean a different throwing forces due to the different weights)

Dick
Homework Helper
At the point of flying off the platform the platform is no longer exerting a force on the objects, but the objects and the platform are travelling with the same velocity and acceleration. Since gravity is now the only force acting on the objects their acceleration is g downward. Hence the acceleration of the platform is also g downward. That's when they 'fly off'. Nothing to do with the mass.

I Just re-read the question; I didn't realize they were on the spring on the same time

On first read I got the impression that the objects were tested on the spring separately hence my confusion

Dick