Colliding again
On a horizontal frictionless circular track of radius R, lie two small masses m and M, free to slide on the track. Between the two masses is squeezed a spring which, however, is not fastened to m adnd M. The two masses are held together by a string.
If the string breaks, the compressed spring shoots off the two masses in opposite directions, spring itself is left behind. The balls collide when they meet again on the track.
Where does this collision take place?
Expressed your answer as the angular displacement of M.
Your answer will be of the form a(pi)/b
Find the value of a*b
DETAILS :
1- M=300 kg
2- m=250 kg
3- a, b are coprime natural numbers
The answer is 110.
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1 solution
The Most Imporant Thing In The Problem Is To Identify That The Motion Is In Horizontal Plane (I Spent
An Hour Thinking About That Some Information Is Missing Because I Was Analysing It In Vertical Plane).
Well The Solution To The Problem Is Simple
Although It Should Have Been Mentioned That The Length Of Compressed Spring Is Incomparable To The
Circumference Of The Track I Am Assuming This Thing To Be True.
As We See There No External Force In Horizontal Direction On The System Consisting Of The Two Masses And Circular Track.
So We Can Conserve Linear Momentum In Horizontal Direction.
Lets The Say The Velocities Of M And m are v1 and v2 in opposite directions.
So We Have
Mv1 = mv2
For The Masses To Collide The Total Angle Covered By Both Of Them Should Be 360 .(Then Only They Will
Be Able to complete the full track)
Let They Meet after time t .
(v1+v2)t = 2 pi R
t = 2 pi r / v1+v2
Now Angular Displacement Of Mass M = v1*t/R
Putting The Value Of Time In Above Equation We Obtain
Angular Displacement = 2*pi / (1+v2/v1)
Now v2/v1 = M/m = 300/250 = 6/5
Angular Displacement = 10*pi/11
Therefore a= 10, b=11