Final kinetic energy of 2kg and 8kg masses after perfectly inelastic collision

  • Thread starter Thread starter hodgepodge
  • Start date Start date
  • Tags Tags
    Energy Final
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
11 replies · 4K views
hodgepodge
Messages
47
Reaction score
0

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
 
Physics news on Phys.org
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
 
v = 8.9443 m/s right... so
2 x 8.443? + 8 x 0 = 2 + 8 x vf

But everything else seems good.