A ball of mass .44 kg and speed of 4.5 m/s collides head-on with a .22 kg ball at rest. Assuming perfect elasticity, what is the speed and direction of each ball afterwards?
I know that the momentum is conserved here and since I am solving fo two variables, I think I would get
.44(4.5) + .22(0) = .44v'1 + .22v'2
transform the equation to solve for v'1 and plug v'1 into the original equation to solve for v'2 then solve again for v'1 actual? when i do this I get 1 for both and this is incorrect so i am very confused please help?
No.
Take that equation and solve for v'1 in terms of v'2.
Then write theconservation of energy equation, putting in v'1 above. Solve.
i think that's exactly what i did and it didn't come out right
let me see you work.
everything cancels out and becomes 6.68 = 6.68. do you mean ke = pe or 1/2mv^2 = 1/2mv'^2. i am really confused
I mean the conservation of energy.
KE original= KE final
in the original, only the m1 was moving. In the final, both have some KE.
so how do i go about solving this
let me see your work. I am not going to do it for you.
what if i said please?
(.5(.66)(4.5)^2) = (.5(.22)Vi^2)+(1/2(.44)(sqroot(6.68-.11Vi^2)/.22)^2
everything cancels out, and the sq root part is derived from solving for Vf
6.68=.11Vi^2 + .22(6.68-.11Vi^2/.22)
6.68=.11Vi^2 + 6.68 - .11Vi^2
6.68 = 6.68
No. Here is the error:
(.5(.66)(4.5)^2) = (.5(.22)Vi^2)+(1/2(.44)( V2^2)/
I have no idea how you got what you did. From the momentum equation, you should have
.44(4.5) + .22(0) = .44v'1 + .22v'2
or v'2=.44(4.5-1)/.22
NOW NOTE that v'2 is the velocity of v2 after the collision, that 2 is not a square. PUt this into..
(.5(.66)(4.5)^2) = (.5(.22)Vi^2)+(1/2(.44)( V2^2)
or
(.5(.66)(4.5)^2) = (.5(.22)Vi^2)+(1/2(.44)( (.44(4.5-1)/.22)^2)
Now do the algebra.
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