Solving Momentum Problems: Cart Interactions and Impulses

  • Thread starter beckster
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In summary: Thus, the force is the same for both carts.The time, however, is different. It took longer for the spring to push cart B because it moved further. The carts have different acceleration vectors, and thus different velocities, but the forces acting on them are the same. This means that the time it took for the carts to hit the bumpers must have been different.Thus, the impulse on cart A was less than the impulse on cart B. The ratio of the impulses applied to the carts is 1:2.In summary, the carts were initially at rest and held together by a string. When the string was burned, the spring expanded and the carts moved apart. Cart A moved 0.45m while cart B moved
  • #1
beckster
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1. Homework Statement [/b]

Two heavy frictionless carts are at rest. They are held together by a loop of string. A light spring is compressed between them. When the string is burned, the spring expands from 2 cm to 3 cm and the carts move apart. Both hit the bumpers fixed to the table at the same instant but cart A moved 0.45 m while cart B moved 0.87m. What is the ratio of:
a) the speed of A to that of B after the interaction?
b) their masses?
c) the impulses applied to the carts?
d) the acceleration of the carts while the spring pushes them apart?


Homework Equations



mava+ mbvb= mava' + mbvb'

1/2 mv2= FfΔd


The Attempt at a Solution


First you will have to figure out everything you know about the carts before the string is burnt. You would then use the other part of the equation to solve after the string is burnt.
You would use the second equation that is shown by filling in the knowns.
 
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  • #2
beckster said:
1. Homework Statement [/b]

Two heavy frictionless carts are at rest. They are held together by a loop of string. A light spring is compressed between them. When the string is burned, the spring expands from 2 cm to 3 cm and the carts move apart. Both hit the bumpers fixed to the table at the same instant but cart A moved 0.45 m while cart B moved 0.87m. What is the ratio of:
a) the speed of A to that of B after the interaction?
b) their masses?
c) the impulses applied to the carts?
d) the acceleration of the carts while the spring pushes them apart?


Homework Equations



mava+ mbvb= mava' + mbvb'

1/2 mv2= FfΔd


The Attempt at a Solution


First you will have to figure out everything you know about the carts before the string is burnt. You would then use the other part of the equation to solve after the string is burnt.
You would use the second equation that is shown by filling in the knowns.

Part c should be easy.

Impulse = force x time

We don't know how strong the force was, and we don't know how long it lasted for, but with our understanding of Newton's Third Law, we do know that the spring was pushing on each cart with the same sized force at any instant.
 

Related to Solving Momentum Problems: Cart Interactions and Impulses

What is momentum and how is it calculated?

Momentum is a measure of an object's motion, taking into account both its mass and velocity. It is calculated by multiplying an object's mass by its velocity.

How do you solve momentum problems involving cart interactions?

To solve momentum problems involving cart interactions, you must use the conservation of momentum principle. This states that in a closed system, the total momentum before an interaction is equal to the total momentum after the interaction.

What is an impulse and how is it related to momentum?

An impulse is a force applied over a period of time, causing a change in an object's momentum. It is equal to the change in momentum and is calculated by multiplying the force by the time it is applied.

What are some common misconceptions about solving momentum problems?

One common misconception is that an object's momentum can be changed by simply changing its velocity. In reality, both mass and velocity play a role in determining an object's momentum. Additionally, some may think that the conservation of momentum principle only applies to objects moving in a straight line, when it actually applies to all interactions between objects.

What are some real-life applications of momentum problems and impulse?

Momentum problems and impulse are important in understanding and predicting the motion of objects in various scenarios, such as collisions in sports, car accidents, and rocket propulsion. They are also crucial in engineering and designing efficient and safe systems, such as airbags in cars.

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