# Homework Help: Basic momentum/loss of energy question

1. Oct 25, 2012

### oreosama

1. The problem statement, all variables and given/known data
Two skaters collide and grab onto each other on frictionless ice. one of them of mass "m1" is moving to the right at "v1", while the other, of mass "m2", is moving to left at "v2"

given m1, m2, v1, v2, determine:

the magnitude and direction of the velocity just after the collision
the loss in kinetic energy of the system

2. Relevant equations

p=mv

3. The attempt at a solution

m1*v1 - m2*v2 = (m1+m2)v3

v3 = (m1*v1-m2*v2) / (m1+m2)

firstly I should know if that was right.. then:

1/2*m1*v1^2 + 1/2*m2*v2^2 = 1/2(m1+m2)(v3)^2

m1*v1^2 + m2*v2^2 = (m1+m2)(m1*v1-m2*v2)^2 /(m1+m2)^2

m1*v1^2 + m2*v2^2 = (m1*v1-m2*v2)^2 /(m1+m2)

(m1*v1^2 + m2*v2^2) (m1+m2) = (m1*v1-m2*v2) (m1*v1-m2*v2)

m1^2*v1^2 + m1*m2*v1^2 + m1*m2*v2^2 + m2*v2^2 = m1^2*v1^2 - 2*m1*m2*v1*v2 + m2^2*v2^2

m1*m2*v1^2 + m1*m2*v2^2 = -2*m1*m2*v1*v2
v1^2 + v2^2 = -2*m1*m2*v1*v2 / ( m1*m2)

2*v1*v2 + v1^2 - v2^2

theres a chance I have a clue I knew what I'm doing on the first part.. the second part clearly not?

2. Oct 25, 2012

### tiny-tim

hi oreosama!
yes
the question asks for the difference in KE …

you should have a minus in the middle of those lines!

now just gather the v12 terms, the v22 terms, and the v1v2 terms​

3. Oct 25, 2012

### oreosama

thanks for that

peddling along through this assignment i've come across:

A bomb of mass "m" at rest explodes. Half of the mass is thrown off in the +x-direction at a speed "v". A quarter of the mass is thrown off in the +y-direction at a speed "3v".

given[m,v]

determine the magnitude and direction of the remaining piece

determine the energy of the explosion

i have no idea what I'm supposed to do, but I would think everything should cancel out?

(1/2*m*v) i + (1/4*m*3v)i + something = 0

-(1/2*m*v) i - (3/4*m*v) j = something ?

everything begins at rest and blows up..

(1/2*1/2*m*v^2 + 1/2*1/4*m*(3v)^2) * 2 (the third piece being exact opposite of the first two means energy should be double the first 2 added up?)

...

11/4*m*v^2

4. Oct 26, 2012

### tiny-tim

hi oreosama!

(just got up :zzz:)
(try using the X2 button just above the Reply box )

yes, momentum (and angular momentum) is conserved in every collision

but your equation should be (1/2*m*v) i + (1/4*m*3v)j + something = 0, shouldn't it?

try again