Basic Impulse Problem

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In summary, a 70 kg boy jumps off a 50 kg cart moving at 2.0 m/s on a frictionless floor, causing him to hit the floor with zero velocity. Using the equation I= mv-mv(initial), the impulse given to the cart is calculated to be -40 kgm/s. The total momentum of the cart and boy before the jump is 120 kgm/s, with the momentum of the cart separately being 100 kgm/s. The momentum of the boy after the jump is 0 kgm/s. The momentum of the cart after the jump can be represented as 50v, where v is the unknown velocity. The total momentum remains 120 kgm/s, with 100 kg
  • #1

Homework Statement



A 50. kg cart is moving across a frictionless floor at 2.0 m/s. A 70. kg boy, riding in the cart,
jumps off so that he hits the floor with zero velocity.

a. What impulse did the boy give to the cart?
b. What was the velocity of the cart after the boy jumped?

Homework Equations



I= mv-mv(initial)

The Attempt at a Solution



(50kg)(2.0m/s)-(120kg)(2.0m/s)=-40 kgm/s

I really have no idea how to get the velocity after he jumped out so I'm pretty sure the impulse is incorrect as well. Any help would greatly be appreciated. Thanks.
 
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  • #2
Answer these questions:

1) What was the total momentum of cart and boy before the boy jumped? What was the momentum of the cart separately?

2) What was the momentum of the boy after he jumped?

3) Calling the velocity of the cart after he jumped 'v', what was the momentum of the cart after he jumped?

4) How does the total momentum in 1) compare with the sum of the momenta in 2) and 3)? Can you solve for v?

5) How much did the momentum of the cart change? That's the impulse.

Then you're done.
 
  • #3




Hello! Thank you for providing the problem and your attempt at a solution. Let's take a closer look at the situation and see if we can come up with a correct solution.

First, let's define what impulse is. Impulse is the change in momentum of an object and is calculated by multiplying the force applied to an object by the time it is applied for. So, in this case, the boy jumping off the cart is the force applied, and the time it takes for him to jump off is the time it is applied for.

a. To find the impulse the boy gives to the cart, we need to calculate the change in momentum of the cart. Since the boy jumps off with a velocity of 0 m/s, the final momentum of the cart will also be 0 kgm/s.

Initial momentum of the cart = (50 kg)(2.0 m/s) = 100 kgm/s
Final momentum of the cart = 0 kgm/s

Therefore, the change in momentum of the cart is -100 kgm/s. This means that the boy gives an impulse of 100 kgm/s to the cart.

b. To find the velocity of the cart after the boy jumps off, we can use the conservation of momentum principle. This states that the total momentum before an event is equal to the total momentum after the event.

Total momentum before = Total momentum after
(50 kg)(2.0 m/s) + (70 kg)(0 m/s) = (50 kg)(v) + (70 kg)(0 m/s)
100 kgm/s = (50 kg)(v)
v = 2.0 m/s

Therefore, the velocity of the cart after the boy jumps off is still 2.0 m/s. This makes sense because there is no external force acting on the cart, so its momentum should remain constant.

I hope this helps clarify the problem and provides a correct solution. Keep up the good work!
 

1. What is a basic impulse problem?

A basic impulse problem is a physics problem that involves calculating the change in momentum of an object after a given impulse is applied to it. An impulse is a force applied over a short period of time, and is typically represented by the symbol "J".

2. What is the equation for calculating impulse?

The equation for calculating impulse is J = Δp, where J is the impulse in Newton-seconds (N∙s) and Δp is the change in momentum in kilogram-meters per second (kg∙m/s). This equation can also be written as J = FΔt, where F is the average force applied over a given time interval, Δt.

3. How is the principle of conservation of momentum related to basic impulse problems?

The principle of conservation of momentum states that in a closed system, the total momentum before an event is equal to the total momentum after the event. In basic impulse problems, this means that the initial momentum of an object before an impulse is applied will be equal to the final momentum after the impulse is applied.

4. What are some real-world examples of basic impulse problems?

Some real-world examples of basic impulse problems include a golf club hitting a golf ball, a tennis racket hitting a tennis ball, a person jumping off a diving board, or a rocket launching into space. In each of these scenarios, an impulse is applied to an object, causing a change in its momentum.

5. How can basic impulse problems be applied in engineering and technology?

Basic impulse problems are commonly used in engineering and technology to design structures and machines that can withstand or control impulsive forces. For example, engineers may use basic impulse calculations to design car airbags, shock absorbers, or crumple zones in order to reduce the impact of an impulse during a car crash. In technology, basic impulse problems can be applied in the design of rockets, missiles, or spacecraft, to ensure precise control of their motion and direction.

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