Question

We are studying elastic collisions. There is this system of equations problem where we are solving for the first and second final velocities. Help!

equations: m1v1,i+m2v2i = m1v1f+m2v2f
0.5m1v1i^2+0.5m2v2i^2 = 0.5m1v1f^2 + 0.5m1v1f^2

Givens: m1= 2kg
m2=3kg
v1,i= 4m/s
v2,i=-5m/s
unknown:
v1,f= ?
v2,f=?
Please help!

Answer 1

we can plug in

Answer 2

Then I simplified one side and used the substitution technique

Answer 3

yes that looks correct

Answer 4

what you did, you might have solved for v_2_f

Answer 5

I have the same expression but the 2 and 3 are swapped. Can you please solve this problem?

Answer 6

$$
\Large
{
m_1 v_{1_i} + m_2 v_{2_i} = m_1 v_{1_f} + m_2 v_{2_f} \\

0.5~m_1v_{1_i}^{2}+0.5~m_2v_{2_i}^2 = 0.5~m_1v_{1_f}^2+0.5~m_2v_{2_f}^2
\\ \therefore \\
2(4) + 3(-5) = 2( v_{1_f} ) + 3(v_{2_f})
\\0.5 (2)(4^{2}) + 0.5 (3) (-5)^{2} = 0.5 (2) v_{1_f}^{2} + 0.5(3) v_{2_f}^{2}

\\ \therefore \\
-7 = 2( v_{1_f} ) + 3(v_{2_f})
\\53.5 = 1~ v_{1_f}^{2} + 2.5~ v_{2_f}^{2}
\\ \therefore \\
-7 - 3(v_{2_f}) = 2( v_{1_f} )\\
\frac{-7 - 3(v_{2_f})}{2} = v_{1_f}
\\ \therefore \\

53.5 =1\cdot \left ( \frac{-7 - 3(v_{2_f})}{2} \right)^2 + 2.5~ v_{2_f}^{2}


}
$$

Answer 7

why are the 2 and 3 swapped?

Answer 8

I probably plugged them in incorrectly.

Answer 9

m1 = 2, m2 = 3 . i might have made a mistake as well, so double check that

Answer 10

m1 = 2kg , m2 = 3kg
v_1_i = 4 m/s , v_2_i = -5 m/s

Answer 11

ok so far?

Answer 12

Yes

Answer 13

where did the four in the last line come from?

Answer 14

$$
\Large
{
m_1 v_{1_i} + m_2 v_{2_i} = m_1 v_{1_f} + m_2 v_{2_f} \\

0.5~m_1v_{1_i}^{2}+0.5~m_2v_{2_i}^2 = 0.5~m_1v_{1_f}^2+0.5~m_2v_{2_f}^2
\\ \therefore \\
2(4) + 3(-5) = 2( v_{1_f} ) + 3(v_{2_f})
\\0.5 (2)(4^{2}) + 0.5 (3) (-5)^{2} = 0.5 (2) v_{1_f}^{2} + 0.5(3) v_{2_f}^{2}

\\ \therefore \\
-7 = 2( v_{1_f} ) + 3(v_{2_f})
\\53.5 = 1~ v_{1_f}^{2} + 2.5~ v_{2_f}^{2}
\\ \therefore \\
-7 - 3(v_{2_f}) = 2( v_{1_f} )\\
\frac{-7 - 3(v_{2_f})}{2} = v_{1_f}
\\ \therefore \\

53.5 =1\cdot \left ( \frac{-7 - 3(v_{2_f})}{2} \right)^2 + 2.5~ v_{2_f}^{2}

\\ \therefore \\

53.5 =\frac{\left ( -7 - 3(v_{2_f})\right)^2}{2^2} + 2.5~ v_{2_f}^{2}
\\ \therefore \\
\color{red} {2^2} \cdot 53.5 = \color{red} {2^2}\left( \frac{\left ( -7 - 3(v_{2_f})\right)^2}{2^2} + 2.5~ v_{2_f}^{2} \right )
\\ \therefore \\
2^2\cdot 53.5 =\left ( -7 - 3(v_{2_f})\right)^2 + 2^2\cdot 2.5~ v_{2_f}^{2}\\
\\ \therefore \\
4\cdot 53.5 = (-7)^2 + 2(-7)(-3)v_{2_f} + 9v_{2_f}^2 + 4\cdot 2.5~ v_{2_f}^{2}

}
$$

Answer 15

okay I see

Answer 16

lets solve this and then use the link to verify if it works

Answer 17

never mind

Answer 18

Also this equation is found here
http://en.wikipedia.org/wiki/Elastic_collision#One-dimensional_Newtonian

it is easier to label u1 , u2 for initial speeds. v1,v2 for final speeds

Answer 19

what did you get?

Answer 20

Vi= -6.8m/s I think

Answer 21

yes! wolfram alpha had similar results

let me check the other value

Answer 22

no

Answer 23

is the link ok

Answer 24

yes it is corrected here:
http://www.wolframalpha.com/input/?i=solve+2*4+%2B+3*%28-5%29+%3D+2x+%2B+3y+%2C+1%2F2+*+2*4^2+%2B+1%2F2+*+3*%28-5%29^2+%3D+1%2F2+*+2*x^2+%2B+1%2F2+*+3*y^2

Answer 25

sorry i made a typo when i put in the equation. its fixed now

Answer 26

It is totally fine. I got 2.4 m/s for v2 but the wolfram alpha calculation says 3.7

Answer 27

so let me go back and simplify this equation, see if we can solve

Answer 28

$$\Large {
2\cdot 4 + 3\cdot (-5) = 2x + 3y \\
\frac{1}{2} \cdot 2\cdot 4^2 + \frac12\cdot 3(-5)^2 = \frac12 2 \cdot x^2 + \frac12 3 \cdot y^2




}
$$

Answer 29

multiply both sides of that second equation by 2, to remove the 1/2

Answer 30

$$

\Large {
2\cdot 4 + 3\cdot (-5) = 2x + 3y \\
\frac{1}{2} \cdot 2\cdot 4^2 + \frac12\cdot 3(-5)^2 = \frac12 2 \cdot x^2 + \frac12 3 \cdot y^2
\\ \iff \\
2\cdot 4 + 3\cdot (-5) = 2x + 3y \\
2\cdot 4^2 + 3(-5)^2 = 2 \cdot x^2 + 3 \cdot y^2
\\ \iff \\
-7 = 2x + 3y\\
107 = 2x^2 + 3y^2
\\ \therefore \\

107 = 2x^2 + 3 \left( \frac{-7-2x}{3}\right )^2
\\ \iff \\
107 = 2x^2 + 3 \cdot \frac{(7+2x)^2}{3^2} \\
\\107 = 2x^2 + \frac{(7+2x)^2}{3} \\

\\ \iff \text {multiply both sides by 3 } \\
\\3\cdot 107 = 3\cdot 2x^2 + (7+2x)^2\\ \iff \\
\\ 321 = 6x^2 + 7^2 + 28x + 4x^2 \\ \iff \\
\\ 321 = 10x^2 + 28x + 49
\\ \iff \\
\\ 0 = 10x^2 + 28x -272

\\ \iff \\
\\ 0 = 5x^2 + 14x -136
\\ \iff \\
0 = ( 5x + 34) (x - 4)
\\ \iff \\
x = -34/5 , 4


}

$$

Answer 31

It is better to solve the problem generally and use the formula derived, than to solve this equation for specific values.
also you can reject the solution x = 4, y = -5 since that implies there was no collision

Answer 32

right