Calculating Mass of Cart After Transfer of Energy with a Running Boy

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Homework Help Overview

The problem involves a scenario where a boy jumps onto a cart, and the discussion centers around calculating the mass of the cart using principles of momentum. The boy has a mass of 50 kg and runs at a speed of 10.0 m/s before jumping onto the cart, which is initially at rest. After the boy jumps on, the combined system moves at a velocity of 2.50 m/s.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the conservation of momentum as a relevant principle and question the definitions of initial and final momentum. There are various attempts to set up equations relating to the mass of the cart and the velocities involved.

Discussion Status

Participants are actively engaging with the problem, exploring different equations and interpretations of momentum. Some have suggested specific equations to use, while others are clarifying the relationships between the variables involved. There is a mix of correct and incorrect reasoning, with no clear consensus on the final answer yet.

Contextual Notes

Some participants express uncertainty about their algebraic manipulations and the correct application of momentum principles. There are indications of confusion regarding the setup of the equations and the roles of the variables involved.

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A 50 kg boy runs at a speed of 10.0 m/s and jumps onto a cart originally at rest. The cart, withe the boy on it, then takes off in the same direction in which the boy was running. If the cart with the boy has a velocity of 2.50 m/s what is the mass of the cart?

I need the equation to find the Mass of the cart. Please help:cry:
 
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Pizall said:
A 50 kg boy runs at a speed of 10.0 m/s and jumps onto a cart originally at rest. The cart, withe the boy on it, then takes off in the same direction in which the boy was running. If the cart with the boy has a velocity of 2.50 m/s what is the mass of the cart?

I need the equation to find the Mass of the cart. Please help:cry:
What physical principle do you think applies to this situation?
 
Pizall said:
A 50 kg boy runs at a speed of 10.0 m/s and jumps onto a cart originally at rest. The cart, withe the boy on it, then takes off in the same direction in which the boy was running. If the cart with the boy has a velocity of 2.50 m/s what is the mass of the cart?

I need the equation to find the Mass of the cart. Please help:cry:
What is the definition of momentum?

Think conservation of momentum. What is the initial momentum of the boy and cart? What is final momentum of boy and cart after he jumps on?
 
I have the equation that starts

m1v1 = m1v1' + m2v2'
 
I first work it like m2v2 = m1v1 but not sure it that was right
 
Pizall said:
I have the equation that starts

m1v1 = m1v1' + m2v2'
It is actually m1v1 +m2v2 = m1v1' + m2v2'. but here, v2 is 0, so you're equation is sort of OK. But what can you say about v1' and v2'? How does each relate to the given v_f = 2.5?
 
v1' is initial velocity which equals 0, and v2' is the the combonation of the boy and the cart velocity which equals 2.50
 
Pizall said:
v1' is initial velocity which equals 0, and v2' is the the combonation of the boy and the cart velocity which equals 2.50
No, its total initial momentum = total final momentum. v1 is the initial velocity of the boy(10). v2 is the initial velocity of the cart(0). v1' is the final velocity of the boy, and v2' is the final velocity of the cart. The boy and the cart move together after he jumps on it. So what's v1' and v2'?
 
Oh ok I think I am with you, the both are 2.50 m/s
 
  • #10
so would you work it out like

m1v1 - m1v1'/ v2' =m2
 
  • #11
Pizall said:
Oh ok I think I am with you, the both are 2.50 m/s
Yes, now you just have to solve for m2. Note that since the boy and cart move together at the same speed and direction after he jumps on it, you could have written the equation as

m1v1 +m2v2 = (m1 +m2)v_f
50(10) + 0 = (50 +m2)(2.5)
and get the same result for m2.

m2 =?
 
  • #12
so m2 = 3 kg

hopfully that is right
 
  • #13
Pizall said:
so m2 = 3 kg

hopfully that is right
You had the right equation, but slipped up on the parentheses and the algebra.
m2 = (m1v1 - m1v1')/ v2'
m2 = ((50)(10) - 50(2.5))/2.5
m2 = (500 - 125)/2.5
m2 = 375/2.5
m2 = ?
 
  • #14
ok m2 = 150 kg right?

I did come up with 150 kg the second time I work it but was not sure.
so I worked it a different way and came up with 3 kg, what I did was cancel out m1 on each side, but now I see that I can't do that because I didnot have m1 only but also had m2 so i could not cancel out m1

I had (v1-v1')/ v2 = m2 I guess that is not right.
 
  • #15
Pizall said:
ok m2 = 150 kg right?

I did come up with 150 kg the second time I work it but was not sure.
so I worked it a different way and came up with 3 kg, what I did was cancel out m1 on each side, but now I see that I can't do that because I didnot have m1 only but also had m2 so i could not cancel out m1

I had (v1-v1')/ v2 = m2 I guess that is not right.
The 150kg mass for m2 is correct. you could have reworked your equation to read
m1(v1 -v1')/v2' = m2, from which,
50(v1 - v1')/v2' = m2
50(10 - 2.5)/2.5 = m2
50(7.5)/2.5 = m2
150 = m2
m2 = 150kg
In doing the algebra, I often find it easier to plug in as many numerical values into the original equation first, before trying to solve for the unknown, because it is easier to work with numbers than it is with letters.
So if you have
m1v1 + m2v2 = m1v1' + m2v2', and you wish to solve for m2, rather than first isolating m2 by itself and easily getting mixed up witht the letters, put in the values first, thus
50(10) + m2(0) = 50(2.5) + m2(2.5)
500 + 0 = 125 + 2.5m2
375 = 2.5m2
150 = m2
if u see what i mean.
 

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