Question

Collisions Lab
http://phet.colorado.edu/sims/collision-lab/collision-lab_en.html
Run the simulation and draw a diagram that represents Pi - the momentum before the collision.
Underneath that diagram draw another diagram that shows two vectors - P1f and P2f - the final momentums of ball 1 and ball 2 after the collision. Include a third vector that represents Pf - the sum of the former two vectors.
Using formula, calculate the total kinetic energy before and the total energy after the collision.
Repeat step 1 but first change the Elasticity to 50%.
Reset the elasticity to

Answer 1

HInt:
if the collision is elastic perfectly, then kinetic energy is conserved and also total momentum is conserved

Answer 2

so, we can write these equations:
\[\Large \begin{gathered}
{m_1}v = {m_1}{u_1} + {m_2}{u_2} \hfill \\
\hfill \\
\frac{1}{2}{m_1}{v^2} = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2 \hfill \\
\end{gathered} \]
where \(v=1,\;m_1=0.5,\;m_2=1.5\), furthermore, \(u_1,\;u_2\), are the corresponding speeds after collision of the two objects

Answer 3

please solve such system for \(u_1,\;u_2\)

Answer 4

what about the two diagrams? how do u draw them?

Answer 5

which formula is for after collision and which one for before?

Answer 6

in this formula:
\[\frac{1}{2}{m_1}{v^2} = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}{m_2}u_2^2\]
left side represents the initial kinetic energy, namely before collision, and right side represents the kinetic energy after collision

Answer 7

im confused how to solve for u1 and u2

Answer 8

and how do u draw the diagrams?

Answer 9

from your data we can write this: \(m_2=3m_1\). Am I right?

Answer 10

\(m_2=1.5\), and \(3m_1=3 \cdot 0.5=1.5\)

Answer 11

so we can rewrite such system as below:
\[\left\{ \begin{gathered}
{m_1}v = {m_1}{u_1} + 3{m_1}{u_2} \hfill \\
\hfill \\
\frac{1}{2}{m_1}{v^2} = \frac{1}{2}{m_1}u_1^2 + \frac{1}{2}3{m_1}u_2^2 \hfill \\
\end{gathered} \right.\]
Now I cancel \(m_1\) from both equations:
\[\left\{ \begin{gathered}
v = {u_1} + 3{u_2} \hfill \\
\hfill \\
\frac{1}{2}{v^2} = \frac{1}{2}u_1^2 + \frac{1}{2}3u_2^2 \hfill \\
\end{gathered} \right.\]
then I multiply both sides of the second equation, by \(2\), so I get:
\[\left\{ \begin{gathered}
v = {u_1} + 3{u_2} \hfill \\
\hfill \\
{v^2} = u_1^2 + 3u_2^2 \hfill \\
\end{gathered} \right.\]
the last system is easily solved using the substitution method. Please try

Answer 12

sorry i dont know how to..:/

Answer 13

from first equation I get \(u_1\) as follows:
\[{u_1} = v - 3{u_2}\]
then I substitute into the second equation:
\[{v^2} = {\left( {v - 3{u_2}} \right)^2} + 3u_2^2\]
which is a quadratic equation, please try to solve it

Answer 14

oh sorry i accidentally tried doing elimination oops

Answer 15

hint:
after a simplification, I get:
\[{v^2} = {v^2} + 9u_2^2 - 6v{u_2} + 3u_2^2\]

Answer 16

this is soo confusing..isnt there another way to solve for kinetic energy?

Answer 17

can you first help me with drawing the diagrams?