How Does Impulse Affect Cart and Boy Dynamics?

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SUMMARY

The discussion focuses on a physics problem involving a 50 kg cart moving at 2.0 m/s and a 70 kg boy who jumps off the cart, landing with zero velocity. The impulse given to the cart by the boy is calculated using the formula I = mv - mv(initial), resulting in -40 kg·m/s. To find the cart's velocity after the boy jumps, the conservation of momentum principle is applied, leading to the conclusion that the total momentum before the jump equals the total momentum after the jump, allowing for the calculation of the cart's new velocity.

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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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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.
 

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