# Inelastic collisions

1. Dec 7, 2009

### jamesdubya

1. The problem statement, all variables and given/known data

A mass weighing 1kg at the top of a frictionless hill where theta = 50 and is 3m long is released from rest and collides inelastically at bottom with a
second mass weighing .25 at rest. The pair then slides up the second incline thats .5m high and fly’s off with a certain velocity V.
Find:
a) the kinetic energy of m1 just
before the collision.
b) the kinetic energy and velocity
of the pair just after the
collision
c) how much kinetic energy was
lost in the collision
d) the velocity of the pair as they fly off the second incline.
2. Relevant equations
ke=1/2mv^2 i think?
pe=mgh
m1v1+m2v2=m1v1f+m2v2f

3. The attempt at a solution
Cannot figure it out, every answer i get is way off, just would like to see how and what equations are used please. Thank you i really appreciate it. Like i said in my other thread im sorry for not trying to show work but i am really really clueless at this point

2. Dec 7, 2009

### denverdoc

Remember that even if energy isn't conserved, mo is. They stick together so recast your last equation with a common velocity, ie

m1v1+m2v2=(m1+m2)vf and since m2 is at rest: we have m1v1= (m1+m2)vf where Vf is the velocity headed back up the hill. We will need to compute another velocity after this one.

Now v1 is not constant it accelerates down the hill so right before collison what is v1: it is no different than if it simply fell vertically through the same vertical displacement, which is what.

sin 50= delta y/ 3 meters

Now find an eqn that relates delta y to velocity. Hint it can be found by considering that the potential energy of mass 1 is traded for kinetic energy.

3. Dec 7, 2009

### jamesdubya

So would it be something like Sin(50) = 1/2mv2 / 3? im not sure i understand what your saying, thank you for your help i appreciate it.

4. Dec 7, 2009

### denverdoc

One step at a time delta y=3 * sin(50).

Now use this delta y (vertcal displacement) in an energy eqn:

mg(delta y)=1/2mv^2 so yes, you are almost right. Now compute the velocity of the two masses stuck together and compute the kinetic energy of the aggregate.

5. Dec 7, 2009

### jamesdubya

1(9.81)(-.7871245611)=.5(1)v^2/3
-46.33015167=v^2
is that right?
Im kinda confused on what to do next

6. Dec 7, 2009

### denverdoc

Not sure where the negative sign is coming from. sin(50)=0.766 then Mult by 3=2.30, That is the fall down the first slope. Notice that m falls out so
gy=1/2v^2 Compute v. This is the speed of the first block. Now it collides with block 2, use the eqn I gave in the first post to compute what I called Vf. This is the speed of the two blocks together. If you have their total mass and their velocity, you can compute kinetic energy of the aggregate. (Hint it should be smaller than the potential energy of the first block).

7. Dec 7, 2009

### denverdoc

BTW you are using way more precision than is justified or needed in your calculations, round to two decimal places for all answers.

8. Dec 7, 2009

### jamesdubya

My calculator was in radians whoops :) well after I fixed everything I got 4.75 J but the answer on the study guide says 35 J so i must be doing somethin wrong. Thank you for your patience as you can tell im not to quick with this stuff

9. Dec 7, 2009

### denverdoc

2.3*1*9.8 is the potential energy=??? Not equal to 4.75 but neither is it 35.

I wasn't sure what was meant by long, the length of the slope (hypotenuse or the base)--do you have a pic?
It must be that the drop is 3,57 meters. So my mistake, use 3.57 m for delta y

in other words tan 50=y/3

10. Dec 7, 2009

### jamesdubya

Sorry, by long i ment going across the x axis http://img121.imageshack.us/img121/1195/41421378.png [Broken]

Last edited by a moderator: May 4, 2017
11. Dec 7, 2009

### denverdoc

So compute V1 using PE=35 J

Then use that velocity as follows: Vf(1.0 kg + 0.25kg)=V1(1.0 kg)

12. Dec 7, 2009

### denverdoc

Very well, we are ok. In your work show y=3m(tan(50)=3.57m

EDIT: Note that 3.57*9.81*1=35.0 J

13. Dec 7, 2009

### jamesdubya

then i do 3.57*9.81 anddd... got it! thank you so much :) so then i would do
m1v1= (m1+m2)vf
right?which ends up beign 35/1.25 which is the correct answer 28J. So what do i do to find the velocity after the collision? thank u alot for your help

14. Dec 7, 2009

### denverdoc

Wait, 28J looks good but do you see where it comes from?

The mass is 5/4 of the original mass, but the velocity is squared and is 4/5 o the original

so the two effects work together to get 28J. Just want to make sure you understand this point as you haven't formally gone thru the 1/2m Vf^2 numerical work.

Assuming you get this we are almost home. We use the same principle of energy conservation to compute the final velocity as it goes off the ramp.

At top of ramp PE+KE=the KE at the bottom of the hill=28J

So how do we get PE for the two blocks at the top of the ramp?

15. Dec 7, 2009

### denverdoc

THIS is Vf in the equation above--maybe not the best term as we have a final final velocity yet to get.

16. Dec 7, 2009

### jamesdubya

is it just 1(9.81)(.5) + .5(1.25)v^2=28?

17. Dec 7, 2009

### denverdoc

The potential energy (you really need to memorize some of these equations) is mgh

m in this case is the two masses stuck together, 1.25Kg
the h is the height of the hill (what we called y in the first half)
g is gravitational constant.

Because some of the kinetic energy at the bottom is lost to PE as we go uphill, the final KE is less than 28.

It is less by the amount of potential energy gained.

It (KE) is computed by 1/2 m v^2 where m is again 1.25kg. We just need to compute the new v and we are done!!!!

18. Dec 7, 2009

### jamesdubya

thank you so much! i really appreciate it, you made my first thread here a good one :)

19. Dec 7, 2009

### denverdoc

Remember Potential energy is that gained by position in the universe--maybe positional energy is a better term for now and kinetic energy is that associated with movement.

So we start with potential energy of block 1 which is converted to kinetic energy which collides with another block in an inelastic collison, and the two have a new total lesser kinetic energy (28J), some of which is lost as the two gain a new vertical position.