Calculating Speed of a Rolling Cart with Added Mass and Running Man

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Homework Help Overview

The problem involves a 1000 kg cart rolling at 1.70 m/s with a 70.0 kg man on it, who begins to run to the left at 7.00 m/s relative to the cart. Participants are exploring how to calculate the new speed of the cart after the man starts running.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and question the definitions of the masses involved. There are attempts to set up equations to relate the initial and final velocities of the cart and the man.

Discussion Status

Some participants have provided equations and reasoning based on conservation of momentum, while others are questioning the initial conditions and calculations. There is no explicit consensus on the correct approach or final outcome, but multiple interpretations and calculations are being explored.

Contextual Notes

There are discussions around the initial momentum calculations and whether the mass of the man should be included in the initial momentum of the system. Participants are also considering the implications of the man's movement relative to the cart.

ElectricMile
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A 1000 kg cart is rolling to the right at 1.70m/s . A 70.0 kg man is standing on the right end of the cart.What is the speed of the cart if the man suddenly starts running to the left with a speed of 7.00 m/s relative to the cart?


really boggles my mind, should i be using

=(m1 +m2)Vf - m1(Vi)/m2
or what's going on?!
 
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(m1+m2)vb=m1*v1a+m2*v2a
we put the positive direction as right
(m1+m2)vb/m1-m2*v2a/m1=v1a

Im not sure how you define the masses but i think you missed (m1 +m2)Vf/!m2!
 
Conservation of momentum. Initially, the 1000kg cart is rolling to the rightare 1.7 m/s while the man is standing still, so the total momentum is (mass)(velocity) 1700 kgm/s.

Let v be the speed of the cart after the man starts running. When the man (mass 70.0 kg) runs to the right at 7.00 m/s, relative to the cart, his speed relative to the ground is v- 7 and so his momentum, relative to the ground, is 70(v- 7)= 70v- 490 kgm/s. The momentum of the cart is 1000v and so the total momentum is now
1070v- 490 . By conservation of momentum, that must be
1070v- 490= 1700=> 1070v= 2190 so v= 2190/1070= 2.05 m/s to the right.
 
Isn't the initial momentum a result of the mass of both the cart and the man, 1070 x1.7 or 1819 to the right ? If so, leads to 2.15 m/s
 
regor60 said:
Isn't the initial momentum a result of the mass of both the cart and the man, 1070 x1.7 or 1819 to the right ?
Yes, the initial momentum must include the man moving with the cart.
 

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