Problem About Change in Momentum/Impulse

  • #1

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 both carts forward for a distance of 1 m, starting from rest.

After traveling 1 m, is the momentum of the plastic cart greater than, less than, or equal to the momentum of the lead cart?

Homework Equations


##J = \Delta p = Favg \times \Delta t##
##\Delta x = (\Delta p \times \Delta t)/ m##
(from the equation for impulse and change in momentum: \Delta x / \Delta t = v_f because the carts start at rest)
##\Delta x = v_i \times \Delta t + .5 \times a_x \times \Delta t^2##
##\Delta x = .5 \times (v_i + v_f) \times \Delta t##

The Attempt at a Solution


I did algebra using delta x and tried both kinematics equations. The entire reason I posted this is because, using both kinematics equations, I derived the statement ##\Delta p = .5 \times \Delta p##. So I must have made a mistake somewhere but I can't figure out where. The acceleration is constant so I should be able to use the kinematics equations.

I think there are solutions to this problem on the Internet, but the main reason I posted this is because of the momentum equals half of itself solutions I got. But if someone wants to help me with the solution to the actual problem as well, that would be appreciated. I messed around with equations and tried to figure out the behavior of the two carts compared to each other, but wasn't yet able to find a solution to how they compare at 1 m.

edit: sorry for the messed up Latex before this edit, i had an appointment to go to so i didn't have time to fix it before i left after i saw it was wrong
 
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Answers and Replies

  • #2
haruspex
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##J = \Delta p = F_avg x \Delta t##
Please do not use x for multiplication in LaTeX. Use \times.
##\Delta x = (\Delta p \times \Delta t)/ m##
mΔx/Δt will give average momentum, not change in momentum.
\Delta x / \Delta t = v_f because the carts start at rest)
Δx/Δt does not give final velocity.
 
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  • #3
Please do not use x for multiplication in LaTeX. Use \times.

mΔx/Δt will give average momentum, not change in momentum.

Δx/Δt does not give final velocity.
Please do not use x for multiplication in LaTeX. Use \times.

mΔx/Δt will give average momentum, not change in momentum.

Δx/Δt does not give final velocity.
Oops! I see my mistake now. Thanks!
 
  • #4
Okay, I figured it out.

I managed to derive that the momentum will be inversely proportional to the final velocity, so since the smaller cart has a bigger velocity its final momentum will be less.

But I also looked at another physics forum post and saw that you can apply the version of Newton's second law that says the definite integral of the force over a time interval is equal to the rate of change of momentum over that interval. So, since the time interval for the faster accelerating cart is shorter, its integral is smaller, thus the change in momentum is less! That's the best way to answer this problem. Cool!
 
  • #5
SammyS
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Okay, I figured it out.

I managed to derive that the momentum will be inversely proportional to the final velocity, so since the smaller cart has a bigger velocity its final momentum will be less.
The above statement makes me want to say, "Whoa there! Momentum is given as mv thus it's proportional to velocity."

Yes, I know that in this case the numbers work out so that ratio of velocities (small : large) is 10 : 1 and the ratio of momenta is 1 : 10 .

The analysis you give below is significantly better.

But I also looked at another physics forum post and saw that you can apply the version of Newton's second law that says the definite integral of the force over a time interval is equal to the rate of change of momentum over that interval. So, since the time interval for the faster accelerating cart is shorter, its integral is smaller, thus the change in momentum is less! That's the best way to answer this problem. Cool!
 
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  • #6
haruspex
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definite integral of the force over a time interval is equal to the rate of change of momentum over that interval.
You are conflating two forms of statement there, leading to an extra level of integration/differentiation.
The definite integral of the force over a time interval is equal to the change of momentum over that interval.
The force at an instant is equal to the rate of change of momentum.
 
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  • #7
You are conflating two forms of statement there, leading to an extra level of integration/differentiation.
The definite integral of the force over a time interval is equal to the change of momentum over that interval.
The force at an instant is equal to the rate of change of momentum.
Thank you for clarifying! That actually came up during class today.

Am I right then that dp/dt would be equal to (N*s)/s then?
 
  • #8
SammyS
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Thank you for clarifying! That actually came up during class today.

Am I right then that dp/dt would be equal to (N*s)/s then?
Yes, assuming that you are referring to units. Of course, 1 Newton*second/second is 1 Newton .

Newton's 2nd Law is often given briefly as ##\ F=ma\ ## or ##\ \displaystyle a = \frac F m \,.\ ## But, it is sometimes given as ##\ \displaystyle F = \frac {dp}{dt} \,,\ ## which I've been told, is the form in which Newton presented it.
 
  • #9
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Quantitatively:

## p_1=Ft_1 ##

and

##p_2=Ft_1\sqrt{\frac{m_2}{m_1}}##
 

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