Calculating Final Energy of Stuck-Together Masses

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Homework Help Overview

The discussion revolves around a problem involving the conservation of momentum and kinetic energy in an inelastic collision between two masses. The original poster presents a scenario where a moving mass collides with a stationary mass, and the goal is to calculate the final energy of the combined masses after they stick together.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss finding the initial velocity of the moving mass using kinetic energy and then applying the conservation of momentum to determine the final velocity of the combined masses. There are questions about the definitions and implications of inelastic collisions, as well as the formulas to use for energy calculations.

Discussion Status

Several participants have provided insights into the calculations and reasoning behind the conservation of momentum and energy. There is ongoing exploration of the correct approach to find the final energy, with some participants verifying their calculations and others questioning the steps taken. The discussion reflects a collaborative effort to clarify concepts and ensure understanding.

Contextual Notes

Participants are navigating the implications of inelastic collisions, including energy loss during the collision process. There is a focus on ensuring that the correct formulas are applied and that assumptions about the system are clearly understood.

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Homework Statement



A 2kg mass with KE=80 J strikes and sticks to an initially stationary 8kg mass. Calculate the final energy of the stuck-together masses.

Homework Equations


KE=1/2 x m x v squared


The Attempt at a Solution


i have no idea, any help would be appreciated
 
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Find the velocity of the 2 kg mass. then apply the law of conservation of momentum. Find the velocity of combined mass and hence the energy.
 
found the velocity, but wat is the law of conservation of momentum? p=mv? if so, i then used K=(p squared)/(2m or m+m) and got 32 J

is that right?
 
Last edited:
p=mv where p is the linear momentum

the conservation of linear momentum says
7745116605c54295c6c3b696cea2d39f.png

where
u signifies vector velocity before the collision
v signifies vector velocity after the collision.

For an inelastic collision
4df879733089c570fbd48698e428fb34.png
 
Last edited by a moderator:
so since i have no epsillon it is inelastic? so i use the formula for inelastic collisions or p=mv?
 
Last edited:
whats your reasoning behind doing that?

Think about it like this, the first mass has kinetic energy initially.
a38c32f3f00f593c1dc17692bc224c0f.png

now it strikes the second mass and they "stick together"... so in this process of sticking energy is lost to heat, bonding etc... but the momentum is conserved since there isn't a net external force on the system.
After they're are stuck they(m1+m2) start moving at a new velocity.
The question asks for the ENERGY in the final mass (m1+m2) after the collision

IF they stick together what kind of a collision is it? elastic or inelastic?
 
inelastic, so i used that formula, and got final velocity of the entire mass to be 1.6886, then i put it into K=1/2 x m(total mass) x v(velocity of total mass) squared and fot 14.2568, am i even close?
 
you're close... can you write down exactly what you're doing so I know where you're going wrong
 
K = 1/2 x m x v squared
80=.5 x 2 x v squared
v = 8.9443 m/s

m1 x v1i + m2 x v2i = (m1+m2)vf
2 x 8.443 + 8 x 0 = 2 + 8 x vf
16.886 = 10 x vf
vf = 1.6886 m/s

K = 1/2 x m x v squared
K = .5 x 10 x (1.6886) squared
K = 14.2568 J
 
  • #10
v = 8.9443 m/s right... so
2 x 8.443? + 8 x 0 = 2 + 8 x vf

But everything else seems good.
 
  • #11
got 16.0001 J

Thank you so much!
 
  • #12
No problem... anytime
 

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