Question

is someone good at physics that can help with this collision question?

Answer 1

@jon91d

Answer 2

vf1/vfi = {(m1-m2)/(m1+m2)}
This is the answer since you have now crossed out v1i (initial velocity of mass 1 is v0 or abbreviated as vknot)
Now, do the same with the second equation.
You JUST have to cross out v0 i.e the initial velocity . thats all!

Answer 3

\[vf1/vfi = {(m1-m2)/(m1+m2)}\]

Answer 4

ok, but what does the 14m2 have to do...where do i plug it in?

Answer 5

m1iv1+m2v2=m1v1f+m2v2f
m1iv1+0=.....()substitute v1f and v2f
Use this law...
Law of conservation of impulse.

Answer 6

\[v_{1f} = (\frac{m_1-m_2}{m_1+m_2})v_0\]\[v_{1i} = v_0\]
\[\frac{v_{1f}}{v_{1i}} = \frac{1}{v_{1i}}*(\frac{m_1-m_2}{m_1+m_2})v_0 = \frac{1}{v_0}(\frac{m_1-m_2}{m_1+m_2})v_0=\frac{m_1-m_2}{m_1+m_2}\]\[m_1=16m_2\]\[\frac{v_{1f}}{v_{1i}} = \frac{16m_2-m_2}{16m_2+m_2} = \frac{15m_2}{17m_2} = \frac{15}{17}\approx0.882\]

You should be able to do the other one with this as a pattern.

Answer 7

As a thought experiment, look at what happens if the two masses are equal. That's what you more typically encounter with billiard balls such as are apparently shown in the diagram. However, if you rolled a lead ball across the table and struck a billiard ball, the lead ball would keep rolling (as we see here, keeping 88% of its original speed if it has 16 times the mass of the ball it strikes) and the stationary ball would go off at a much higher speed than the heavy ball initially had.

Answer 8

you sir, or (ma'am) are a genius! thank you for the thorough clarification. @whpalmer4

Answer 9

Glad I could help!