Momentum question involving two objects of varying mass

In summary: You summed it up perfectly. The time interval of force application on the lighter plastic cart is shorter than that for the heavier lead cart. Using the impulse-momentum theorem, we have ΔP = ∫F dt for the two carts. Since the time interval is smaller for the plastic cart, the definite integral of the force will be smaller resulting in a smaller change in momentum for the plastic cart.In summary, the plastic cart, being lighter in mass, will have a smaller change in momentum compared to the lead cart when the same force is applied for the same distance. This is because the time interval of force application is shorter for the plastic cart due to its higher acceleration.
  • #1
goraemon
67
4

Homework Statement



A 0.2 kg plastic cart and a 20 kg 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? After traveling 1 m, is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart? Explain.

Homework Equations


p = mv
Δs = (Vi+Vf) / 2 * (Vf-Vi) / a
F = ma

The Attempt at a Solution



Using kinematics, I found that the acceleration rate of each cart equals Vf^2 / 2, where Vf equals the final velocity of each cart.

Since Fnet = ma, I find that Fnet of the plastic cart = Fnet(c1) = 0.2kg * Vf(c1)^2 / 2 = Vf(c1)^2 / 10.
Fnet of the lead cart = Fnet(c2) = 20kg * Vf(c2)^2 / 2 = 10 * Vf(c2)^2

Since the problem tells us that the forces are equal for both carts, Fnet(c1) = Fnet(c2).
In other words, Vf(c1)^2 / 10 = 10 * Vf(c2)^2

Solving for Vf(c1), I find that Vf(c1) = 10*Vf(c2)

Since momentum P = mv, P(c1) = (0.2kg)(10*Vf(c2)) = 2 * Vf(c2).
P(c2) = 20kg*Vf(c2)

So I find that P(c1) = P(c2) / 10. Meaning P(c1) < P(c2).

There is no way to check my answer since there's no answer key that I can find and I'm self-studying. Is my answer above correct? Did I go wrong anywhere in my thinking? Thanks a lot in advance.
 
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  • #2
Hello, goraemon. Welcome to PF!

Your work looks correct.

If you have studied the concept of impulse and also the impulse-momentum theorem, then you can use them to get the answer more quickly.

[The work-energy theorem is another approach.]
 
  • #3
Thanks for your response TSny. I do recall reading about the impulse-momentum theorem...change in momentum = definite integral of the Force over some time interval, right?

So, let me see if I got this...since the plastic cart is lighter in mass than the lead cart, the same force exerted on the plastic cart will cause it to accelerate much faster, meaning it will cover the same distance in less time.

So this will result in a shorter time interval for the application of the force, meaning the definite integral of the force will be smaller...therefore, the change in momentum will likewise be smaller for the plastic cart. Is this the right reasoning to use?
 
  • #4
goraemon said:
So, let me see if I got this...since the plastic cart is lighter in mass than the lead cart, the same force exerted on the plastic cart will cause it to accelerate much faster, meaning it will cover the same distance in less time.

So this will result in a shorter time interval for the application of the force, meaning the definite integral of the force will be smaller...therefore, the change in momentum will likewise be smaller for the plastic cart. Is this the right reasoning to use?

Beautiful! :smile:
 
  • #5


Your approach to the problem is correct. Your reasoning and calculations are also correct. You have correctly found that the momentum of the plastic cart is 1/10th of the momentum of the lead cart. This is because momentum is directly proportional to mass and velocity, and the lead cart has 100 times the mass of the plastic cart. Therefore, even though both carts have the same final velocity, the lead cart has a much higher momentum due to its higher mass. This also makes intuitive sense, as a heavier object will have more resistance to changes in its motion and thus will have a greater momentum after traveling the same distance with the same force.
 

Related to Momentum question involving two objects of varying mass

1. How is momentum calculated for two objects of varying mass?

The formula for calculating momentum is mass times velocity (p = mv). For two objects of varying mass, you would calculate the momentum of each individual object and then add them together to find the total momentum.

2. Does the mass of an object affect its momentum?

Yes, the mass of an object directly affects its momentum. The larger the mass, the greater the momentum. This is because momentum is a product of both mass and velocity, and a larger mass requires more force to accelerate to a certain velocity.

3. How does the velocity of an object affect its momentum?

The velocity of an object has a direct effect on its momentum. The greater the velocity, the greater the momentum. This is because momentum is a product of both mass and velocity, and a higher velocity requires more force to accelerate to.

4. What happens to the momentum when two objects of varying mass collide?

In a collision between two objects of varying mass, the total momentum of the system remains constant. This means that the momentum of the individual objects may change, but the total momentum before and after the collision will be the same.

5. Can the momentum of two objects of varying mass be equal?

Yes, the momentum of two objects of varying mass can be equal if their mass and velocity are balanced. For example, if one object has a large mass and a low velocity, while the other has a smaller mass and a higher velocity, their momentums can be equal.

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