Fraction of the initial kinetic energy is transferred?

[DrMcDreamy]

Homework Statement

This is a 2 part question and I already solved for part I. Part II is what I don't get.

a) A 3.16 g object moving to the right at 24.3 cm/s makes an elastic head-on-collision with a 6.32 g object that is initially at rest. The acceleration of gravity is 9.8 m/s2 . 24.3 cm/s 3.16 g 6.32 g Find the velocity of the first object immediately after the collision. Answer in units of cm/s. answer: -8.1 cm/s

b) What fraction of the initial kinetic energy is transferred to the 6.32 g object?

Homework Equations

Would I use: KE=\frac{mv²}{2} ?

The Attempt at a Solution

KE=\frac{(6.32 g)(24.3 cm/s)²}{2} = 1866

We never did this in lecture so I'm not sure if I am right or wrong.

 

[tiny-tim]

Hi DrMcDreamy!

(in LaTeX, you must use ^ not SUP )

Yes, kinetic energy (of a non-rotating body) is always mv²/2

(though in this case, I think you've mixed up the m of one body with the v of another)

 

[DrMcDreamy]

Hi DrMcDreamy!

(in LaTeX, you must use ^ not SUP )

Yes, kinetic energy (of a non-rotating body) is always mv²/2

(though in this case, I think you've mixed up the m of one body with the v of another)

Im new , thank you. So it would be:

KE= \frac{(6.32 g)(-8.1 cm/s)^2}{2}

since I want the fraction of the intial kinetic energy of that is transferred to the 6.32 g object?

 

[tiny-tim]

Hi DrMcDreamy!

KE= \frac{(6.32 g)(-8.1 cm/s)^2}{2}

since I want the fraction of the intial kinetic energy of that is transferred to the 6.32 g object?

I'm getting confused

The 6.32 g object is stationary beforehand, and after the collision the other object has a speed of 8.1 cm/s.

So the other object's https://www.physicsforums.com/library.php?do=view_item&itemid=132" after the collision is 3.16*(8.1)²/2 ergs, and before the collision it was 3.16*(24.3)²/2 ergs, which is 9 times as much …

so what fraction of the initial kinetic energy is transferred to the 6.32 g object?

 

[DrMcDreamy]

Hi DrMcDreamy!

I'm getting confused

The 6.32 g object is stationary beforehand, and after the collision the other object has a speed of 8.1 cm/s.

So the other object's https://www.physicsforums.com/library.php?do=view_item&itemid=132" after the collision is 3.16*(8.1)²/2 ergs, and before the collision it was 3.16*(24.3)²/2 ergs, which is 9 times as much …

so what fraction of the initial kinetic energy is transferred to the 6.32 g object?

I'm sorry I confused you! I figured it out, Its supposed to be \frac{KEf}{KEi}

Here my work:

 

[tiny-tim]

yes, that's the right result …

(but why didn't you use the 8.1 in the answer to part a) ?)

 

[DrMcDreamy]

yes, that's the right result …

(but why didn't you use the 8.1 in the answer to part a) ?)

Because the -8.1 is the velocity from the first object and its asking for the fraction of the initial kinetic energy that is transferred to the 6.32 g object, which is the second object. The -8.1 cm/s is the final velocity of the first object.

 

[tiny-tim]

Because the -8.1 is the velocity from the first object and its asking for the fraction of the initial kinetic energy that is transferred to the 6.32 g object, which is the second object. The -8.1 cm/s is the final velocity of the first object.

But it would be much easier to use the 8.1 …

the fraction of KE transferred is 1 - the fraction kept, so that's 1 - KE_f/KE_i, both for the first object …

the advantage of this is that since all the parameters are the same, except for the speed, you don't need to do a lot of multiplying …

you just say that the ratio of the speeds is 1/3, so the ratio of the KEs is … ?

 

[bulldogGHS]

How do we solve the first part of the question?

 

[tiny-tim]

welcome to pf!

hi bulldogGHS! welcome to pf!

How do we solve the first part of the question?

use conservation equations …

show us what you get ​

(gravity is irrelevant)