# Stopping distance of two objects with equal Kinetic Energy

I'm working on a problem regarding two objects with the same kinetic energy.

Two objects with masses of $m_1$ and $m_2$ have the same kinetic energy are both moving to the right. The same constant force F is applied to the left to both masses. If $m_1=4m_2$, the ratio of the stopping distance of $m_1$ to that of $m_2$ is:

I believe the key information here is that both objects have the same kinetic energy. This must mean that an equal amount of work (force $\times$ distance) must be applied to each object in order to bring it to a complete stop ($KE + q = 0$)?

Is the correct answer $1:1$?

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Followup question: What is the ratio of the times required to stop each object? – dmckee May 1 '13 at 20:38
Not an answer but check out physics.stackexchange.com/questions/535/… as it may shed some light on this for you. – Brandon Enright May 1 '13 at 21:52

Yes, you have spotted the key fact from the question.

The heavier mas is going 1/2 as fast, and experiences 1/4 the acceleration, which by v^2 = 4ad rearranged as d = v^2/4a, shows that the ratio of stopping distance for both objects will be identical.

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Answer is 1:1. Here is how you can derive it

Since kinetic energy is equal

1/2 (m1*v1^2)=1/2 (m2*v2^2) => 4m2 *v1^2 = m2*v2^2 taking root on both sides

2v1 = v2

Since Force is equal, deacceleration a1 = F/m1 and a2= F/m2

Now use v^2 = u^2-2as for both and derive s1:s2 = 1:1

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