Elastic and inelastic collisions help

Homework Statement

here is a picture of the problems im having trouble with:
http://oi55.tinypic.com/2llca50.jpg
or if that doesn't work: http://tinypic.com/view.php?pic=2llca50&s=7

Homework Equations

for question 1:
part 1: p1 + p2 = p1 + p2 and Distance=Time*Speed
part 2: W(f) = F(f) x d, where F(f) is the Frictional force and d is the distance moved, and F(f) = μN, where μ is the coefficient of friction (i think)
part 3: i don't understand how there is a height acheived, other than falling 5m down the ramp

for question 2:
part 1, 2 and 3: Ke=1/2mv^2 AND m1v1+m2v2 = m1'v1'+m2'v2'

The Attempt at a Solution

question 1:
part 1: im really not sure how to figure in friction to this part, all i really know is mgh plays a part. but to what extent, im not sure. this is my biggest problem.

part 2: since i don't know the distance moved, i can't find out the work done by friction

part 3: -5m is the only height achieved that i could see as plausible

question 2:
part 1: 9.6 + 0 = 3.2v+2.0v
v = 9.6/3.2/2.0/2 = .75m/s
Ke = 1/2 (3.2+2)(.75)^2 = 1.46J

part 2: v = .75m/s+(9.8m/s*.2857s[calculated using an online calculator]) = 3.54m/s
Ke = 1/2 (5.2)(3.54)^2 = 32.5j

part 3: theyre different because the connected boxes and gained velocity due to gravity during a freefall, which caused a rise in kinetic energy

i'm fairly clueless as to whether i got any of that right, and you can see i need a strong push in the right direction on the first one.

Last edited:

so i did some more more and here's what i got

question 1
part 1.) a. 16.7m b. 4.91
part 2.) a. 3.34 b. .982
part 3.) i still dont understand at all

question 2
1.) 8.4J
2.) 29J
3.) potential energy becomes kinetic energy during the fall etc etc

haha u go to UofM borther?

Mishra's class? was just about to post the same thing!!!

gneill
Mentor
The diagram indicates an initial height of 5m, but the problem text states 2.5m. Which is correct?

For part 3, since the block sliding down the ramp collides with a stationary block of twice the mass, you can expect the less massive block to recoil (bounce back to some degree). Hence it will travel back up the ramp to some height that depends upon the recoil velocity.

The diagram indicates an initial height of 5m, but the problem text states 2.5m. Which is correct?

For part 3, since the block sliding down the ramp collides with a stationary block of twice the mass, you can expect the less massive block to recoil (bounce back to some degree). Hence it will travel back up the ramp to some height that depends upon the recoil velocity.

its 2.5m, sorry about that. a hack job in paint

and so for part 3, would i use 1/2 mVo^2= mgh ?

gneill
Mentor
its 2.5m, sorry about that. a hack job in paint

and so for part 3, would i use 1/2 mVo^2= mgh ?

Yes, you'll have to determine the recoil velocity of m1 and then apply the above to find h.

Yes, you'll have to determine the recoil velocity of m1 and then apply the above to find h.

How would you find the work though?

gneill
Mentor
How would you find the work though?

Are you referring to the work referenced in part 2? At first glance, it looks to me as though the best you can do is find expressions for the work in terms of the mass, m1. The work done in slowing the bodies from whatever speed down to zero will equal their initial kinetic energies, or force times distance. You would need to know the actual mass of the object in order to determine the KE or the force (due to friction).