Elastic and inelastic collisions help

In summary: 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).
  • #1
duxerwhere
3
0

Homework Statement


here is a picture of the problems I am 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 = p`1 + p`2 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: I am really not sure how to figure in friction to this part, all i really know is mgh plays a part. but to what extent, I am 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.
i really appreciate your help
 
Last edited:
Physics news on Phys.org
  • #2
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 don't understand at all

question 2
1.) 8.4J
2.) 29J
3.) potential energy becomes kinetic energy during the fall etc etc
 
  • #3
haha u go to UofM borther?

Mishra's class? was just about to post the same thing!
 
  • #4
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.
 
  • #5
gneill said:
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 ?
 
  • #6
duxerwhere said:
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.
 
  • #7
gneill said:
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?
 
  • #8
BugMeNot_dude said:
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).
 

What is an elastic collision?

An elastic collision is a collision between two objects in which no kinetic energy is lost. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision.

What is an inelastic collision?

An inelastic collision is a collision between two objects in which some kinetic energy is lost. This means that the total kinetic energy of the system before the collision is greater than the total kinetic energy after the collision.

How is momentum conserved in a collision?

Momentum is conserved in a collision because the total momentum of the system before the collision is equal to the total momentum after the collision. This means that the sum of the momentum of each object in the system remains the same.

What is the difference between elastic and inelastic collisions?

The main difference between elastic and inelastic collisions is the conservation of kinetic energy. In an elastic collision, kinetic energy is conserved, while in an inelastic collision, some kinetic energy is lost. This results in different outcomes for the objects involved, such as changes in velocity and deformation.

How can I calculate the final velocities in an elastic collision?

To calculate the final velocities in an elastic collision, you can use the conservation of momentum and conservation of kinetic energy equations. These equations can be solved simultaneously to find the velocities of the objects after the collision. Additionally, you can use the coefficient of restitution, which is a measure of the elasticity of the collision, to calculate the final velocities.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
755
  • Introductory Physics Homework Help
Replies
15
Views
238
  • Introductory Physics Homework Help
Replies
12
Views
530
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
879
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top