Comparing Momentum of Two Carts with Different Masses

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When equal forces are applied to two carts of different masses over the same distance, the lighter cart (0.2 kg) will achieve a higher velocity than the heavier cart (20 kg). However, the momentum of each cart depends on both mass and velocity. Although Cart A has a higher velocity, Cart B's significantly greater mass compensates for this, resulting in a greater overall momentum for Cart B. The work done on both carts is equal, leading to the conclusion that momentum can be compared using the relationship between force, distance, and mass. Understanding these principles clarifies the momentum dynamics between the two carts after the push.
einsteinette
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Hello there, I am having difficulty with this question:

A 0.2 kg plastic cart (Cart a) and a 20 kg (Cart b) lead cart can both roll without friction on a horizontal surface. Equal forces are used to push the two carts forward for a distance of 1 m, starting from rest. Which cart has the greater momentum, after the full 1m?

So I was thinking that since you apply equal forces to push the two carts, Cart A would have a higher velocity. But when you do the momentum of the two carts together, the larger mass of cart B would make up for the difference of velocity. If I'm right about the velocity thing that is.

mava = 0.2Va(>vb)
mbvb = 20Vb(<va)

So the momentum would be somewhat equal?
 
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einsteinette said:
So I was thinking that since you apply equal forces to push the two carts, Cart A would have a higher velocity.
That's true, but you need to figure out how much higher is the velocity of Cart A. Hint: Since equal force is exerted for equal distance, what can you say is the same for the two carts after the push?
 
The force applied and the rate of deceleration?
 
einsteinette said:
The force applied and the rate of deceleration?
The force is the same but the accelerations are not. But you can use Newton's 2nd law and some kinematics to figure out the velocity of each cart after it is pushed. Then you can compare momentum.

But another way is to recognize what force*distance gives you, and use that to figure out the velocity and momentum.
 
Ahh so you mean work done is the same. I think I get it now. Thanks!
 
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