# Physics (mechanics) AP Problem-involving kinetic energy of a moving object

• physicsgirl101
Kinetic energy is not conserved in inelastic collisions.In summary, this is an AP Physics Mechanics question involving a moving railroad car and sand pouring onto it. The problem entails determining the mass and velocity of the car as a function of time, as well as its initial and final kinetic energy. The normal force exerted on the car by the tracks is also calculated. The sand pouring rate is constant, and the sand's momentum is not conserved due to external forces. The collision between the sand and car is inelastic, leading to a decrease in final kinetic energy.
physicsgirl101
Physics (mechanics) AP Problem---involving kinetic energy of a moving object

The following is an AP Physics Mechanics question. I have some questions about understanding what it's asking me to do and about impulse.

PROBLEM:

An open-top railroad car (initially empty and of mass Mo) rolls with negligible friction along a straight horizontal track and passes under the spout of a sand conveyor. When the car is under the conveyor, sand is dispensed from the conveyor in a narrow stream at a steady rate (delta-M)/(delta-t)=C and falls vertically from an average height h above the floor of the railroad car. The car had initial speed Vo and sand is filling it from time t=o to t=T. Express your answers to the following in terms of the given quantities and g.

a) Determine the mass M of the car plus the sand that it catches as a function of time t for 0<t<T.

b) Determine the speed V of the car as a function of time t for 0<t<T.

c) i. Determine the initial kinetic energy Ki of the empty car.
ii. Determine the final kinetic energy Kf of the car and its load.
iii. Is kinetic energy conserved? Explain why or why not.

d) Determine expressions for the normal force exerted on the car by the tracks at the following times.
i. Before t=0
ii. For 0<t<T
iii. After t=T

a)Mass = Mo+Ct
b) V=(MoVo)/(Mo+Ct)
c) i. Ki=(1/2)MoVo
ii. kf = (1/2)(MoVo)
iii. ? (they shouldn't be equal, as I have it, should they?)
d) i. Mg
ii. (Mo+Ct)g
iii. (Mo+Ct)g

I know I didn't get it all right... Alright, here are MY QUESTIONS (regarding specific letters of the answers)---(IF YOU CAN EVEN HELP ME WITH ONE OF THESE QUESTIONS I'D BE SO GRATEFUL!):

a) should the "t" in this answer be lower case or upper case? I was confused about that.
b)I set the initial and final momentums equal to solve this part (b)... But I feel like I never took the sand into account in the initial momentum. I had: MoVo=(Mo+Ct)V (initial momentum = final momentum).
c)shouldn't c) i. and ii. NOT be the same?----because energy isn't conserved in collisions(right?)

another question: I never took the height the sand falls (h) into account anywhere, which I'm sure i should have... in which part---and how?

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Part a): It should be M0 + CT

The method: The mass rate of flow of the sand is constant and given by:

$$\frac{\Delta M}{\Delta t} = C$$

The total change in mass over a time interval is the rate of change of mass with time times the total time interval:

What I just said is:
$$\Delta M = \frac{\Delta M}{\Delta t} \Delta t$$

$$= C (T - 0 ) = CT$$

$$M = M_0 + \Delta M = M_0 + CT$$

Part b)

It seems that you have correctly understood that since there is negligible friction, no external forces act on the rail car. Therefore, momentum is conserved. You have applied conservation of momentum. The momentum is given by M0v0...and the velocity as a function of time is the momentum divided by the mass as a function of time: v(t) = p/m(t) = m0v0/(m0 + Ct).

Yes, this time the t is lowercase...hopefully you understand why. The mass is being expressed as a function of time. t is the variable t that gives you elapsed time. T, on the other hand, is the total time that has elapsed after that sand pouring time interval is over (it is a constant).

Well done!

*Edit* Not sure what to say about the momentum of the sand. Remember it is under the influence of external forces, so its momentum is not conserved. Gravity accelerates it until it hits the bed of the rail car and stops dead...presumably it exerts the same force downward on the car as the car exerts upward on it to stop it. I don't think this has any effect on the problem.

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Part c)

Initial kinetic energy looks right, but wouldn't the final kinetic energy be given by 1/2 (M0 + CT)(vf)^2 ?

And we know from the previous part that vf = m0v0/(m0 + CT)

= [m0/(m0 + CT)]v0...which is less than v0 by the same factor that the new mass is greater than m0. However, this factor is squared for the velocity, so the final kinetic energy decreases overall. Why? I am not totally sure. But the collision between the sand and rail car is definitely inelastic, wouldn't you agree? Kinetic energy is not conserved in inelastic collisions.

d i) and ii) look correct

iii) should be (Mo + CT)g

I honesly haven't taken the sand into consideration either. Momentum is conserved, but all of the momentum the sand gains is vertical. Since it comes to a stop after falling, all of that vertical momentum must be transferred to the rail car. But it has nowhere to go. I'm assuming that the bed of the car would begin vibrating...or the shocks would absorb the impact...or something.

Energy is more tricky. I stated that kinetic energy is not conserved because it is an inelastic collision, but I am not at all sure of myself where or how some of the inital kinetic energy of the truck is lost/dissipated...can anyone elaborate?

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cepheid said:
Part c)

Initial kinetic energy looks right, but wouldn't the final kinetic energy be given by 1/2 (M0 + CT)(vf)^2 ?

And we know from the previous part that vf = m0v0/(m0 + CT)

= [m0/(m0 + CT)]v0...which is less than v0 by the same factor that the new mass is greater than m0. However, this factor is squared for the velocity, so the final kinetic energy decreases overall. Why? I am not totally sure. But the collision between the sand and rail car is definitely inelastic, wouldn't you agree? Kinetic energy is not conserved in inelastic collisions.

Part C) the initial and final kinetic energy is wrong. It should be velocity SQUARED... and besides vf does not equal vo...

As well Part a) is right... Cepheid thought that it was the mass at t = T, however it's actually asking for the mass as a function of t where it is changing, not when it equals T.

To prove that Kf does not = Ki

Kf - Ki should = 0 if conserved

(MoVo/(Mo+C(T)))^2 * 1/2 * (Mo + C(T)) - MoVo^2*1/2

= (Mo^2*Vo^2 - MoVo^2(Mo + C(T)))/2(Mo+C(T))

= Vo^2( Mo^2 - Mo^2 -MoC(T))/(2(Mo+C(T)))

= Vo^2(- Mo(C(T))/(2(Mo+C(T)))

No wwhere in the above quation is any variable 0 so thus, kinetic energy is not conserved.

Last edited:
cepheid said:
Part c)

Initial kinetic energy looks right, but wouldn't the final kinetic energy be given by 1/2 (M0 + CT)(vf)^2 ?

And we know from the previous part that vf = m0v0/(m0 + CT)

= [m0/(m0 + CT)]v0...which is less than v0 by the same factor that the new mass is greater than m0. However, this factor is squared for the velocity, so the final kinetic energy decreases overall. Why? I am not totally sure. But the collision between the sand and rail car is definitely inelastic, wouldn't you agree? Kinetic energy is not conserved in inelastic collisions.

Thanks for all your help! This is totally unrelated but energy IS conserved in elastic collisions, right?

Actually, cepheid, physicsgirl101's answer to a), Mass= M<sub>0</sup>t, was correct. The problem asked for "the mass M of the car plus the sand that it catches as a function of time t for 0<t<T." M<sub>0</sub>T is the mass at the end of that time period.

As for physicsgirl101's last question, yes, kinetic energy is conserved in elastic collisions: that's basically the definition of "elastic collision".

physicsgirl101 said:
Thanks for all your help! This is totally unrelated but energy IS conserved in elastic collisions, right?

Yes, energy is always conserved. physicsgirl101

No one has said much about taking the height the sand falls (h) into account anywhere in the problem, which I'm sure i should have... in which part/in what way does it affect the problem?

(thank everyone so much for all your help so far!), I am very thankful!

## 1. What is the formula for calculating kinetic energy?

The formula for calculating kinetic energy is KE = 1/2 * m * v^2, where m is the mass of the object and v is its velocity.

## 2. Can kinetic energy change?

Yes, kinetic energy can change. It can increase or decrease depending on the change in velocity or mass of the object.

## 3. How is kinetic energy related to work?

Kinetic energy is directly related to work. The work done on an object is equal to the change in its kinetic energy.

## 4. Can an object have a negative kinetic energy?

No, kinetic energy is always positive. If an object is moving in the opposite direction of its velocity, its kinetic energy will be zero.

## 5. How does kinetic energy affect an object's motion?

Kinetic energy is a measure of an object's motion. The greater the kinetic energy, the faster the object is moving and the harder it is to stop.

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