Basic momentum/loss of energy question

[oreosama]

Homework Statement

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

Homework Equations

p=mv

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?

 

[tiny-tim]

hi oreosama!

given m₁, m₂, v₁, v₂, determine:

the magnitude and direction of the velocity just after the collision
v₃ = (m₁*v₁-m₂*v₂) / (m₁+m₂)

firstly I should know if that was right..

yes

the loss in kinetic energy of the system
…
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)

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​

 

[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

 

[tiny-tim]

hi oreosama!

(just got up :zzz:)

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

(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 ​

 

[Nickie]

I would like to point out that the equations and calculations provided in the attempt at a solution are not entirely correct. The initial momentum of the system should be conserved, meaning that the total momentum before the collision should be equal to the total momentum after the collision. This can be represented by the equation m1v1 + m2v2 = (m1 + m2)v3, where v3 is the velocity of the combined skaters after the collision.

To determine the magnitude and direction of the velocity just after the collision, the equation can be rearranged to solve for v3. The magnitude can be found by taking the square root of the sum of the squares of the velocities, and the direction can be determined by considering the signs of the velocities.

The loss in kinetic energy of the system can be calculated by subtracting the final kinetic energy from the initial kinetic energy. The initial kinetic energy is given by 1/2m1v1^2 + 1/2m2v2^2, and the final kinetic energy is given by 1/2(m1 + m2)v3^2. This will give the loss in kinetic energy due to the collision.

It is important to note that in a real-life scenario, there may be some loss of energy due to friction and other factors. This would need to be taken into account in the calculations.