Solving Velocity of Trolley After Loss of Grains

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Homework Help Overview

The problem involves a trolley with a mass of 10 kg carrying grains of mass 5 kg, moving at a velocity of 20 m/s. The grains exit the trolley at a rate of 0.2 kg/s through a hole, and the task is to determine the trolley's velocity after all the grains have fallen out, under the assumption of a frictionless surface.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of conservation of momentum and equations of motion, with some expressing confusion about the relationship between the falling grains and the trolley's velocity.
  • Questions arise regarding the rate at which the grains are lost, with a participant clarifying that it is 0.2 kg/s rather than 0.2 m/s.
  • Some participants question the validity of the provided answer and the underlying assumptions about forces acting on the trolley.
  • There are attempts to derive relationships involving mass and velocity through integration and momentum properties.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have offered insights into the physics involved, while others express uncertainty about the correctness of the provided answer. The conversation reflects a mix of attempts at reasoning and clarifying assumptions.

Contextual Notes

Participants note that the problem was presented by a friend, leading to uncertainty about the accuracy of the answer. There is also a lack of consensus on how the velocity of the trolley changes as the grains fall out, with some questioning the assumptions about external forces.

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Homework Statement



A trolley of mass 10 kg has heap of grains of mass 5 kg in it and it is moving with a velocity of 20m/s. There is a hole made then, so that the grains come out at the rate of 0.2m/s. When all the grains fall out, what is the velocity of the trolley?
(Surface is frictionless, and no grains obstructs motion of trolley as they fall down)

Homework Equations



Equations of motion.

The Attempt at a Solution



So the time I got was
5/0.2 = 25 seconds

then i tried to apply conservation of momentum, but that didnt work.
So I used the equation
v=u+at
v=20+a*25

but since this has 2 unknows, i am not understanding how to solve it.
Any help, appreciated!
 
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The question, as stated, doesn't make sense. Is it really 0.2m/s? Or is it 0.2kg/s? If it's the former, you need either the mass of each grain, or total number of grains (so that you can determine mass of one grain).

Do you know (if this is in your textbook) what the answer is supposed to be?
 
oh sorry, it is 0.2 kg/s
the answer is 10m/s
 
Last edited:
though i never understood how the velocity decreased, i thought it would increase actually.
Any help appreciated!
 
I think your answer book is incorrect. If the grains are falling down (y-direction), and the trolley (with grains) is moving at a constant velocity on a frictionless surface in the x-direction - then there is no force acting in the x-direction to impede/alter the velocity of the trolley.
 
i don't know whether it is right or wrong(the answer). This sum was given to me by a friend and he told me the answer. So are you sure about that, the velocity doesn't change?
 
umm...is the answer really wrong?
 
The legend said:
umm...is the answer really wrong?


I think so.
No external force is acting on the system. So dp/dt = 0.

p = mv

dp/dt = m*dv/dt + v*dm/dt = 0.

dm/dt is constant. Let it be μ.

m*dv/dt = -μ*v.

At any instant m = mo - μ*t.

dv/v = -μ*dt/(mo-μt)

\int_{v_o}^v{\frac{dv}{v}} = \int_0^{25}{\frac{-\mu*dt}{(m_o -\mu*t)}

Solve the integration and find v.
 
Ok, thanks a lot, sir!
 
  • #10
i thought about using momentum properties to solve this, since the variables involve mass and velocity, but idk. maybe the gravitational force times the amount of time it takes for the grain to fall out can equal the final velocity times mass minus the initial velocity times the mass.
 

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