# How Does Momentum and Kinetic Energy Change in a Series of Elastic Collisions?

• PremedBeauty
In summary, the speed of block 3 is greater than the speed of block 1 and the same as the initial values for block 1.
PremedBeauty
block 1 of mass m1 slides along an x-axis on a frictionless floor with speed of v1i=4.00 m/s. Then it undergoes a one dimensional elastic collision with stationary block 2 of mass m2=.500m1. Next, block 2 undergoes a one dimensional elastic collision with stationary block 3 of mass m3=.500m2. (a) What then is the speed of block 3? Are (b) the speed, the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block1?

The lines, the "x" and numbers are kinda croocked, sorry.

a)We use the principle of conservation of momentum here.

Initially momentum is (m1)(4)
After collision with m2, it will be (v)(m1+0.5m1) = (v)(1.5m1)
After collision with m3, it will be (u)(1.5m1+0.25m1) = (u)(1.75m1)
this is bevause m3 = 0.5m2 = 0.5(0.5m1) = 0.25m1.

They must be equal so
(m1)(4) = (u)(1.75m1)
u = 2.3m/s roughly!

(b) Calculate the KE of m3. It is 1/2(2.3)(0.25m1)2
The KE of m1 is 1/2(m1)(42)

See which one is greater!

(d) They are asking the momentum of m3 and not total momentum. Right?This will be (0.25m1)(2.3) where as that of m1 is (m1)(4) so of course that of m1 is greater

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It appears that you are treating the collisions as inelastic collisions. The problem says they are elastic collisions. Both momentum and kinetic energy are conserved. For such collisions there is a velocity difference relationship you might want to use.

Oh, How would you do it if it were elastic?

By definition, in an elastic collision, kinetic energy is conserved. Calculate the total kinetic energy before and after collision.

How do I know which numbers are before and after? Can you show me some how?

PremedBeauty said:
How do I know which numbers are before and after? Can you show me some how?

The momentum conservation equations apply to each collision

m1*v1i = m1*4.00 m/s <== total momentum

m1*v1i = m1*v1f + m2*v2b <== first collision (b subscropt for between collisions

m2*v2b = m2*v2f + m3*v3f <== second collision

If you combine these equations, you have

m1*v1i = m1*v1f + m2*v2f + m3*v3f

The total momentum after both collisions have occurred is the same as the initial total momentum. But this is not enough for you to find the velocities. You need to apply conservation of kinetic energy to each collision. Kinetic energy involves the squares of the velocities, so the algebra can get bit messy. But it is not too difficult to do the calculation algebraically one time to find a relationship between the differences between the velocities of the two masses before and after the collision. Once you have this relationship, you can use it in all elastic collsion problems without dealing with the squares in the energy equation. If your textbeook does not have this equation, see if you can derive it yourself. If you get stuck, look here towrd the bottom of the page.

http://www.astrophysik.uni-kiel.de/~hhaertel/MECHANICS/Mathe/mathe.html

For any 1-dimensional two-mass elastic collision
M1*V1i + M2*V2i = M1*V1f + M2*V2f

M1*V1i² + M2*V2i² = M1*V1f² + M2*V2f²

Solve for V1f - V2f in terms of V1i - V2i

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Hello,

I'm sorry, I'm still not really following. I still have to find the momentum of block B and state is the kinetic, the speeed of block 3 greater than, less than, or same as initial.

PremedBeauty said:
Hello,

I'm sorry, I'm still not really following. I still have to find the momentum of block B and state is the kinetic, the speeed of block 3 greater than, less than, or same as initial.

For any 1-dimensional two-mass elastic collision:

M1*V1i + M2*V2i = M1*V1f + M2*V2f <== Rearrange this momentum equation

M1*V1i - M1*V1f = M2*V2f - M2*V2i <== Factor the Ms

M1(V1i - V1f) = M2(V2f - V2i)

M1*V1i² + M2*V2i² = M1*V1f² + M2*V2f² <== Rearrange this energy equation

M1*V1i² - M1*V1f² = M2*V2f² - M2*V2i² <== Factor the Ms

M1(V1i² - V1f²) = M2(V2f² - V2i²) <== Factor the differences of squares

M1(V1i - V1f)(V1i + V1f) = M2(V2f - V2i)(V2f - V2i)

Divide this last equation by the rearranged factored momentum equation above to get

(V1i + V1f) = (V2f + V2i)

This is usually rearranged and written as

(V1i - V2i) = -(V1f -V2f)

This equation is valid for any elastic collision of two objects in one dimension. Use it with the conservation of momentum equation for each collision in your problem. At least give this a try an post your results.

## 1. What is a collision?

A collision occurs when two objects come into contact with each other, resulting in a forceful impact or interaction between them. In the context of transportation, a collision usually refers to a crash or accident between two or more vehicles.

## 2. How can collisions be prevented?

Collisions can be prevented by following traffic laws and regulations, maintaining a safe distance from other vehicles, avoiding distracted or impaired driving, and regularly maintaining your vehicle to ensure it is in proper working condition. It is also important to stay alert and aware of your surroundings while driving.

## 3. What should I do if I witness a collision?

If you witness a collision, you should immediately pull over to a safe location and call 911 to report the incident and provide any necessary assistance to those involved. It is important to remain calm and follow the instructions of emergency responders.

## 4. What should I do if I am involved in a collision?

If you are involved in a collision, you should first check for any injuries and call for medical assistance if necessary. You should then exchange contact and insurance information with the other driver(s) involved and document the scene with photos or videos if possible. It is also important to report the incident to your insurance company and follow their instructions for filing a claim.

## 5. What are the common causes of collisions?

Some common causes of collisions include distracted driving, speeding, impaired driving, poor weather conditions, and vehicle malfunctions. It is important to always practice safe driving habits and be aware of potential hazards on the road to help prevent collisions.

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