# Problem About Change in Momentum/Impulse

• gibberingmouther
In summary, a 0.2 kg plastic cart and a 20 kg lead cart, both rolling without friction on a horizontal surface, are pushed forward with equal forces for a distance of 1 m, starting from rest. The momentum of the plastic cart is less than the momentum of the lead cart after traveling 1 m. This can be explained by the fact that the momentum is inversely proportional to the final velocity, and since the smaller cart has a bigger velocity, its final momentum will be less. Another way to understand this is by applying the version of Newton's second law that states the definite integral of the force over a time interval is equal to the change of momentum over that interval. Since the time interval for the faster accelerating
gibberingmouther

## 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

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

gibberingmouther
haruspex said:
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.

haruspex said:
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!

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!

gibberingmouther said:
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!

gibberingmouther
gibberingmouther said:
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.

gibberingmouther
haruspex said:
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?

gibberingmouther said:
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.

Quantitatively:

## p_1=Ft_1 ##

and

##p_2=Ft_1\sqrt{\frac{m_2}{m_1}}##

## What is momentum and how does it relate to impulse?

Momentum is a measure of an object's motion, calculated by multiplying its mass by its velocity. Impulse, on the other hand, is the change in an object's momentum over time. It is equal to the force applied to an object multiplied by the time interval in which it acts. In other words, impulse is the force required to change the momentum of an object.

## What factors affect the change in momentum of an object?

The change in an object's momentum depends on the force applied to it and the time interval in which the force acts. The larger the force or the longer the force is applied, the greater the change in momentum will be. Additionally, the mass of the object also plays a role, as a larger mass will have a greater momentum for a given velocity.

## What is the law of conservation of momentum?

The law of conservation of momentum states that in a closed system, the total momentum before an event or interaction will be equal to the total momentum after the event. In other words, momentum is conserved and cannot be created or destroyed, only transferred or transformed.

## What is an elastic collision?

An elastic collision is a type of collision where both momentum and kinetic energy are conserved. This means that the total momentum and total kinetic energy of the system before the collision will be equal to the total momentum and total kinetic energy after the collision. In an elastic collision, the objects involved bounce off each other without any loss of energy.

## How is impulse related to force and time?

Impulse is directly proportional to both force and time. This means that a larger force or a longer time interval will result in a larger impulse. Mathematically, impulse can be calculated by multiplying force by time (I = F x t). This relationship is important in understanding how to minimize the force required to change an object's momentum, such as in sports where padding or extending the time of impact can reduce the force of a collision.

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