How Does Conservation of Momentum Affect Kinetic Energy in Collisions?

• Puddles
In summary: I'm not even sure how to finish that sentence.In summary, a large, moving object will lose more kinetic energy in a collision than a stationary object.
Puddles

Homework Statement

[/B]
(a.)Apply the law of conservation of momentum to the perfectly inelastic collision of a moving object of mass m1 and velocity vi with a stationary object of mass m2
(b.) Solve this for final velocity vf
(c.) Write a formula for the initial kinetic energy (KEi)
(d.) Write a formula for the final kinetic energy (KEf) in terms of vf
(e.) Substitute your formula for vf from (a.) into your equation above to get a formula for KEf in terms of m1, m2, and vi alone
(f.) Find the ratio KEf/KEi in terms of m1 and m2 alone
(g.) Express the final kinetic energy in terms of the initial kinetic energy KEi
(h.) For the case of a moving object much, much more massive than the stationary object (m1 >> m2), how much kinetic energy is lost in the collision?
(i.) For the case of a stationary object much, much more massive than the moving object (m2 >> m1), how much kinetic energy is lost in the collision?
(j.) Given two projectiles fired with the same KE, fired at a target which is free to move, which will cause the most destruction (that is, result in the greatest loss of kinetic energy), a small high-speed projectile or a massive slow-moving one?

Homework Equations

Momentum: P = m x v
Kinetic Energy: KE = 1/2mv^2[/B]

The Attempt at a Solution

(a.) P = M1V1 + M2V2 = (M1+M2)Vf

(b.) Vf = M1V1/(M1+M2)

(c.) Cons. Energy, KEi = KE1 + KE2, KEi = 1/2(M1V1^2) + 1/2(M2V2^2)
KEi = 1/2(M1V1^2)

(d.) 1/2 (M1 + M2) Vf^2

(e.) Here's where my confusion starts, (M1+M2)Vf = M1V1 + M2V2 and 1/2(M1+M2)Vf^2, I tried
1/2(M1+M2)((M1V1+M2V2)/(M1+M2))^2, substituting in for Vf but I'm not sure if that's right

(f.) I'm not even sure how to resolve this from the last piece of information, because I can't set M1 and M1 or M2 and M2 equal to each other from my previous formulas, can I?

Since I'm stuck there I couldn't make an attempt on the rest yet.

Note that, since M2 is initially at rest, V2 = 0.
Hence use this in part (e).

grzz said:
Note that, since M2 is initially at rest, V2 = 0.
Hence use this in part (e).

Okay, so then I've got...
.5(M1 + M2)((M1V1)/(M1 + M2))^2

So then (f.) is...
(.5(M1 + M2)((M1V1)/(M1 + M2))^2) /
(.5(M1V1^2)), which can be simplified, cancel the .5's, a (M1 + M2), and (M1V1)

Okay I got (M1)/(M1 + M2)

How do I use that to get KEf in terms of KEi now? It's KEi times ((M1V1)/(M1 + M2))^2?

Puddles said:
How do I use that to get KEf in terms of KEi now? It's KEi times ((M1V1)/(M1 + M2))^2?

You got
KEf/KEi = M1/(M1 + M2)

hence

KEf = ( M1/(M1 + M2) )KEi.

grzz said:
You got
KEf/KEi = M1/(M1 + M2)

hence

KEf = ( M1/(M1 + M2) )KEi.

Oh, right, thank you. Sometimes I fail to make simple connections in physics, I really appreciate your help.

For (h.) and (i.) then should I just be treating M1 or M2 as zero depending on which part I'm doing because it's got comparatively insignificant mass to the other? So for (h.) for example I would use KEf = (M1(M1 + M2))KEi = (M1(M1 + M2)(.5M1V1^2), where if M1 >> M2, M2 = 0, then compare it to the same formula when M1 = M2? As I'm reading that I know it's not right…

Okay, if I'm trying to find lost KE I should set KEi = KEf and then look at the difference between the two to find what's lost right? Using the idea from above where if one mass is smaller I'm treating it as insignificant? Or I feel like that's no right either, but how do I go about working with one mass as though its smaller than the other without assigning values to the masses?

For part (h),

imagine a very large heavy truck moving and colliding with a very small car. Do you think the initial KE of the truck will decrease much?

grzz said:
For part (h),

imagine a very large heavy truck moving and colliding with a very small car. Do you think the initial KE of the truck will decrease much?

Well no, I'd imagine there would be basically no energy lost, and that the reverse would be true for (i.), but I wasn't sure if it was necessary to show some kind of work behind this thought, maybe just plug in 10^5 vs. 1, or for a question like that is it unnecessary to prove this since it's a simple deduction? I guess that's dependent on my teacher unless it's a point most would consider rudimentary to the point a proof is unnecessary.

Also for (j.) I believe it would be a small-fast moving projectile but I'm not sure if I need to do something to show this.

Puddles said:
Well no, I'd imagine there would be basically no energy lost, and that the reverse would be true for (i.)

You imagined correctly.

I think that as for (h) and (i) you do not have to substitute any numbers. Just consider KEf = ( M1/(M1 + M2) )KEi.

Then ignore the smaller mass when compared to the bigger one in (h) and in (i) and make the obvious deductions as regards what happens to the KEf and KEi.

grzz said:
You imagined correctly.

I think that as for (h) and (i) you do not have to substitute any numbers. Just consider KEf = ( M1/(M1 + M2) )KEi.

Then ignore the smaller mass when compared to the bigger one in (h) and in (i) and make the obvious deductions as regards what happens to the KEf and KEi.

Okay, awesome. Thank you so much for your help! I really appreciate it.

1. What is momentum and why is it important?

Momentum is a physical quantity that describes the motion of an object. It is calculated by multiplying an object's mass by its velocity. Momentum is important because it allows us to predict how an object will behave when it collides with another object or experiences a force.

2. How is momentum conserved in a free response problem?

In a free response momentum problem, the total momentum of the system remains constant. This means that the sum of the momenta of all objects involved in the problem before and after the collision or interaction will be the same. This principle is known as the law of conservation of momentum.

3. What are the units of momentum?

The SI unit for momentum is kilogram meters per second (kg*m/s). However, it can also be expressed in other units such as gram centimeters per second (g*cm/s) or newton seconds (N*s).

4. How does momentum differ from velocity?

Momentum and velocity are related, but they are not the same. Velocity is a vector quantity that describes the speed and direction of an object. Momentum, on the other hand, is a vector quantity that describes an object's motion and its resistance to change in motion. While velocity can change due to external forces, momentum remains constant as long as there is no external force acting on the object.

5. Can momentum be negative?

Yes, momentum can be negative. Since momentum is a vector quantity, it has both magnitude and direction. If an object is moving in the opposite direction of its initial velocity, its momentum will be negative. This can happen in collisions or interactions where the direction of motion changes.

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