Inelastic Collision and kinetic energy?

AI Thread Summary
In an inelastic collision, such as two pieces of taffy colliding and sticking together, momentum is conserved while kinetic energy is not, as some energy is converted into heat. The discussion highlights the need to calculate the kinetic energy before and after the collision to determine the percentage lost to heat. When considering a scenario where one piece of taffy is twice as massive, the final velocity will differ, affecting the kinetic energy calculations. Participants are encouraged to reference their textbooks or online resources to better understand the principles of inelastic collisions and how to derive final velocities from initial conditions. Ultimately, the focus is on understanding the relationship between mass, velocity, and energy loss in these types of collisions.
tvshonk
Messages
3
Reaction score
0

Homework Statement


A piece of taffy slams into and sticks to another identical piece of taffy that is at rest. The momentum of the two pieces stuck together after the collision is the same as it was before the collision, but this is not true of the kinetic energy, which is partly turned into heat. What percentage of the kinetic energy is turned into heat?

(my own addition) What if the initial piece of taffy was twice as massive as the one at rest when it collided?

Homework Equations

The Attempt at a Solution


Something with the equation for kinetic energy? Or is it conservation of momentum because of twice the mass and the same velocity? Not sure how the second relates to loss of energy for friction.
 
Last edited:
Physics news on Phys.org
They're asking about kinetic energy being turned into heat. If some of it is turned into heat then that portion will no longer be contributing to the (bulk) motion of the system.

You have a perfectly inelastic collision. What's conserved? What's the KE before and the KE after collision?
 
Well, the KE isn't conserved since it's lost to heat... but momentum has to be conserved, but I don't see how the equation for that would lead to KE if I just figure out the mass and velocity changes.

If they were the same, but double the mass in the second case... velocity would have to change? And with mv I could find the KE?
 
Last edited:
tvshonk said:
Well, the KE isn't conserved since it's lost to heat... but momentum has to be conserved, but I don't see how the equation for that would lead to KE if I just figure out the mass and velocity changes.
Look in your text, class notes, or on the web to investigate "inelastic collision". How do you determine the final velocity giving the initial masses and velocities of the colliding bodies?
If they were the same, but double the mass in the second case... velocity would have to change? And with mv I could find the KE?
Sure. You know the mass, so if you know mv you can find v, right? What's the expression for KE?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'

Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is（Figure 2）： F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Projectile Compresses a Spring Gun
Replies
10
Views
2K
Potential Energy - 2D inelastic collision - ballistic pendulum
Replies
10
Views
2K
Confused about a conservation of energy problem
Replies
4
Views
2K
Why is momentum conservation always true in a collision?
Replies
4
Views
2K
Kinetic energy and momentum of two satellites colliding
Replies
8
Views
3K
Find elastic energy in the compressed spring.
Replies
12
Views
3K
Elastic Collision and Momentum of Ice Skaters
Replies
16
Views
3K
Question regarding inelastic/elastic collision.
Replies
2
Views
2K
A cylinder rotating on a plane with friction and then moving across a frictionless plane to a collision
Replies
1
Views
757
Kinetic energy and momentum in an elastic collision
Replies
22
Views
3K

Hot Threads

Recent Insights

Back
Top