I understand your answer conceptually, but I'm not sure how to solve it algebraically. Should I start with a collision between M and Mew (which will yield the same velocity) and then Mew with m?
Also, initially I was thinking that m must be:
m = \frac{1}{2} M + \frac{1}{2} Mew
Ah...
I got it?
When a mass M collides with another mass, say M', what is the value of M' such that the whole KE of M is transferred to M' ? Think on these lines.
M transfers all its kinetic energy to M' only if M' is the same mass as M! So the answer is Mew-M's mass must be equivalent to the M's...
Energy is conserved in a completely elastic collision. I used (1) energy of 'M', (2) collision of 'M' and 'm', (3), collision of 'Mew' and 'm', (4) energy of 'm'
First see attached for description of initial height.
So:
(1) Energy
Mg(L-Lcos\theta )=(1/2)MV^{2}
V =...
I am not entirely sure on how to begin and I'm also new to this forum so sorry for the bad formatting.
I have 6 constant acceleration equations by breaking up the following equations into vector equations that correspond to x and y.
Vf = Vo + at
V^{2} = Vo^{2} + 2a(d)
d = do + Vo^{2}...
Lazy Flee!
A lazy flee desires to jump across a log of radius R. It wants to find the most efficient way possible to do so. Find the initial velocity, distance from the radius of the log, and angle so that the above condition is true.
Ok, I need help with a basic collision problem. Please see the attached image and picture the image as a pendulum.
You bring mass 'M' up a certain height and let it go. It collides elastically with mass 'Mew M." Mass 'm' goes up a maximum height. Find mass 'Mew M' so that mass 'm' can attain...