# F = MA Exam 2010 #'s 15,16 (Two mass system with initial velocity)

• SignaturePF
In summary, the student attempted to solve the homework equation for 15, but ran into an issue because the triangular mass was not fixed to the table and was free to slide. He looked to conservation of energy and conservation of momentum to help him solve the equation. When he did, he realized that he needed to account for the final kinetic energy of the triangle. He then applied conservation of energy to solve for the final kinetic energy of the block.
SignaturePF

See:
#15, 16

## Homework Equations

U_o + K_o = U + K
v_CoM = vM1 + vM2 / (M1 + M2)

## The Attempt at a Solution

I thought that for 15 it was simply:
1/2mv^2 = mgh
h = v^2/2g

But this is wrong, so then I looked at the answers and saw that they dealt with CoM. How am I supposed to do this problem? And what should have hinted me toward using the CoM besides the answer choices?

Last edited:
Hi SignaturePF!

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SignaturePF said:
I thought that for 15 it was simply:
1/2mv^2 = mgh
h = v^2/2g

ah, but you're assuming the KE at the top is 0, and it isn't

common-sense (if nothing else! ) tells you that at the top, the velocities of both blocks must be the same

(and as a general rule, conservation of momentum will almost always need to be fitted in somewhere)

show us what you get

When I read the problem I first thought "Conservation of Energy!". Why? Because its easier than kinematic equations and is often the key to a problem. But upon further inspection I quickly came into a issue... The triangular mass is not fixed to the table and is free to slide... This means that I can't equate the initial kinetic energy of 'm' to the final potential energy of 'm'. I have to also consider the final kinetic energy of 'M'. When I write that equation out I have too many variables and can't solve it. To eliminate another variable I appeal to another equation, conservation of momentum. This is the natural thing to do when solving problems like this. Conservation of momentum should be thought of just after (or maybe even before) conservation of energy. Particularly in these types of problems. Go into the problem knowing that you may need to pull one, or both out. Also recall generally if you have one unknown you can solve it with one equation but for two unknowns you need two equations. If you have too many variables and and want to eliminate one (or more) you should think to use conservation of energy or momentum (along with any other pertinent equations of course).

Ok, let's see.
Obviously, as both of you mentioned, it's a perfectly inelastic collision; that is, the two velocities are the same.

mv = (m+M)v'
v' = m/(m+M)v

Now we apply conservation of energy:
U + K = U' + K'
U initial is zero
For K, the triangle and block both start off with v' and it ends up with just the triangle moving while the block eventually stops right?

So:
1/2m2/(m+M)2v2 = mgh + 1/2Mv2
Subtracting right side KE from both sides
(m^2/(M+m) - M)v^2 = 2mgh
Ya this is messed up.

SignaturePF said:
v' = m/(m+M)v

yes

but you haven't applied it correctly here …
1/2m2/(m+M)2v2 = mgh + 1/2Mv2

(and shouldn't the last term should be 1/2mv2 ?)

Ok let me try again. Is the energy thought process correct though.
1/2(M+m)((m/(M+m)v)^2 = mgh + 1/2Mv^2
I think the last term should not be mv^2 because we are looking for when the block has STOPPED. It should be Mv^2 because I thought that the triangle/plane would continue moving.
But simplifying using what you said:
1/2m^2/(M+m)v^2 = mgh + 1/2mv^2
Multiply by 2, divide by m
m/(M+m)v^2 = 2gh + v^2
Subtract v^2:
v^2(m/(M+m) - 1 ) = 2gh
v^2(-M /(M+m) = 2gh - Why is there a negative sign?

h = v^2(M/M+m)/2g
This is the correct answer. But I'm still a bit confused.

SignaturePF said:
I think the last term should not be mv^2 because we are looking for when the block has STOPPED. It should be Mv^2 because I thought that the triangle/plane would continue moving.

your RHS should be the initial KE, which is for the small moving mass, m
Why is there a negative sign?

the final KE must be less than the initial KE,

so yes, you must subtract mgh from the initial KE

I don't think I understand what you're saying. Could you rephrase that?

the initial KE is all you've got

everything else has to come from that

so initial KE = final KE + PE

(and now I'm off to bed :zzz:)

## 1. What is the significance of the "F = MA" equation?

The equation "F = MA" represents Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

## 2. Can you explain the concept of a "two mass system" in this context?

A two mass system refers to a system in which two objects of different masses are connected and moving together. In the context of this exam question, it is likely referring to a pulley system where two masses are connected by a string or rope, and the movement of one mass affects the movement of the other.

## 3. How is initial velocity relevant in this problem?

The initial velocity of the system refers to the velocity of the masses at the beginning of the experiment. This initial velocity can affect the acceleration and final velocity of the system, and is therefore important to consider in the problem.

## 4. What is the purpose of these two questions (15 and 16) on the exam?

These questions are likely testing your understanding of Newton's Second Law and your ability to apply it to a specific scenario. They may also be assessing your ability to analyze and interpret data, as the problem likely involves numerical values for force, mass, and velocity.

## 5. Can you provide any tips for solving these types of problems?

Some tips for solving problems involving Newton's Second Law and two mass systems include drawing free-body diagrams to identify the forces acting on each object, setting up equations using "F = MA" for each object, and using algebra to solve for unknown variables. It is also important to pay attention to the units of measurement and use consistent units throughout the problem.

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