Calculate the final velocity of the cart

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SUMMARY

The final velocity of a cart with three students, each weighing 60 kg, is calculated using the conservation of momentum. The cart has a mass of 120 kg, and when the students jump off at 10 m/s relative to the cart, the final velocity of the entire system with respect to the Earth is determined to be 0 m/s. This conclusion is reached by applying the equation m1v1 + m2v2 = m1v1' + m2v2', where the initial velocities of both the cart and students are zero. The calculation assumes no external forces, such as friction or air resistance, are acting on the system.

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thegreatone09
Three students, each having a mass of 60 kg, climb onto a large flatbed cart that has a mass of 120 kg. Standing at one end and taking turns they run to the opposite end and jump off, one immediately following the other, each with a velocity of 10 m/s with respect to the cart. Calculate the final velocity of the cart and students with resepect to the earth.

I know that we have to use:
m1*v1 + m2*v2 = m1*v1 + m2*v2

the left side is before and the right side is after
m1 = mass of object A
v1 = velocity of object A
m2 = mass of objectB
v2 = velocity of object B

Any help will be appreciated.
 
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well i asssume the system starts at rest, so the left side of your equation would = 0 and your equation would reduce to:
m1*v1 + m2*v2 = 0


m1 = mass of the cart
v1 = final velocity of the cart
m2 = mass of a child
v2 = final velocity of a child

you need another equation to solve for 2 variables:

V1 - v2 = 10

solve it for v1:
v1 = 10 + v2
and plug it into your first equation:

120*(10+v2) + (60)(v2) = 0
v2 = -20/3
plug into the second equation: v1 = 10/3

this is just the first step. you have to do it three times, one for each child, using the result from previous one as the initial for the second.
 


To calculate the final velocity of the cart and students with respect to the Earth, we can use the conservation of momentum equation:

m1v1 + m2v2 = m1v1' + m2v2'

Where m1 and m2 are the masses of the cart and students respectively, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities.

Since the students are jumping off the cart, their final velocities with respect to the Earth will be equal to their initial velocities with respect to the cart (10 m/s). Therefore, we can rewrite the equation as:

(120 kg)(0 m/s) + (60 kg)(0 m/s) = (120 kg + 60 kg)v'

Simplifying, we get:

0 = 180 kg * v'

Since the mass of the cart and students combined is 180 kg, we can see that their final velocity with respect to the Earth will be 0 m/s. This means that the cart and students will come to a complete stop after they jump off.

In conclusion, the final velocity of the cart and students with respect to the Earth is 0 m/s. It is important to note that this calculation assumes that there are no external forces acting on the system, such as friction or air resistance.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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