Why Does a Smaller Mass Reach the End of a Frictionless Track First?

AI Thread Summary
In the discussion, a question arises about two objects of different masses being pushed with equal forces on a frictionless track, leading to confusion about which will reach the end first. It is clarified that the smaller mass will reach the end first due to its greater acceleration, as acceleration is determined by the applied force divided by mass (F=ma). The presence of friction is irrelevant in this scenario since the objects are being actively pushed, resulting in a non-zero acceleration. The relationship between mass and velocity is explained, showing that as mass decreases, velocity increases for a given distance. Ultimately, the smaller mass accelerates faster and therefore reaches the end of the track first.
Q7heng
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Homework Statement


Hi, on my recent test there was a question stating 2 objects, one with a mass of m, and another with a mass of 4m, are pushed with equal forces on a frictionless track. The question asked which object will reach the end first and I said they will reach at the same time because there is no friction. But the right answer is the smaller mass will reach there first. I don't know how this works into everything in the equations of physics and how force, mass and velocity (not acceleration) is related. Help please!

Homework Equations


F=ma

Thanks a lot
 
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You know how force relates to mass and acceleration. Which mass will have the greater acceleration?
How does velocity depend on acceleration?
 
Shouldn't acceleration be 0 in this case because there is no friction on track?
 
Q7heng said:
Shouldn't acceleration be 0 in this case because there is no friction on track?
I have no idea why you would think that.
Acceleration results from a nonzero net force. If there is an applied force and nothing to oppose it then there will be acceleration.
Indeed, there is less likely to be acceleration when there is friction.
 
Q7heng said:
Shouldn't acceleration be 0 in this case because there is no friction on track?

That would be the case if they were coasting down the track. However the question as written in post #1 states they are being pushed down the track by a force.
 
∑Fx=ma (Newton's Second Law)

F(Push)=ma The only force acting in the direction of motion is the push.

We can also write Newton's second law for the y-direction, but it doesn't tell us anything. ∑Fy=N-mg=0 →N=mg.

Back to the x-direction, we found that a=F(Push)/m by solving Newton's second law in the x-direction for a.

Now we can use kinematics. Since distance is implicitly given, I would use the equation,

v2=2a(xf-xi)=2aΔx

So,

v=√2aΔx or v=√2(F/m)Δx

since m is on the bottom, for a given Δx (i.e. the length of the track) v gets larger as m gets smaller
 

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