How Does Impulse Affect Cart and Boy Dynamics?

[youngblood]

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.

 

[Dick]

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.

 

[m2003]

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!