Recoiling Buggy!
A buggy of mass 100 kg is free to move on a frictionless horizontal track. Two men, each of mass 50 kg, are standing on the buggy, which is initially stationary. The men jump off the buggy with velocity of 10 m/sec relative to the buggy. In one situation, the men jump one after the other. In another situation, the men jump simultaneously.
If the ratio of the recoil velocities of the buggy in two cases can be expressed in the form of where and are coprime natural numbers, then calculate the value of .
Details And Assumptions I Dont Think that i need to tell the direction of jumping of the two men . it could be easily inferred from the question
For The Sake Of Simplicity Assume That In Both Situations Direction Of Jumping Of Men Are The Same (Means That If In One Situation They Jump In Same Direction Then In Other situation Also They Will Jump In Same Direction And vice-versa)
The answer is 13.
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1 solution
Lets Analyse Both The Situation One By One
Situation 2
Firstly We Should Note That Men Have To Jump In Same Direction Otherwise
The Buggy Will Not Move At All (By Conserving Momentum Because There Is No
External Force).If Buggy Will Not Move Then v = 0 And In Taking Ratio Of
Velocities Denominator Will Become Zero (As Clearly State a,b are coprime
natural numbers).So taking Zero Does Not Make Any Sense
Now Come To The Problem
If They Jump Simultaneously
Apply Momentum Conservation
Initial Momentum Of System = 0
50 * 10+50 * 10+100 * v = 0
v = -10 m/sec
So Buggy Will Move In Opposite Direction Of Jumping
Situation 1
Again Men Will Jump In Same Direction
Applying Momentum Conservation When Only One Men Jumps off
50 * 10+150 * v = 0
v = -10/3 m/sec
Now One Man And Buggy Will Continue Moving With This Velocity
We Should Note That We Apply Momentum Conservation wrt ground only
so first let us find velocity of second man wrt ground (Because in the problem it is
clearly stated that the men jump of with velocity 10 m/sec wrt buggy in both
situations)
On Applying Concept Of Relative Velocity
We Get
Vm = 40/3 m/sec (Opposite To The Movement of buggy)
Now Apply Momentum Conservation
Please Note That Initial Momentum Is NONZERO in this case
We Get
150 * 10/3 = -50 * 40/3+100 * v
On Simplification We Get
v = 35/3 m/sec
Taking Ratio Of The Results Obtained In Both Situations We Get
v1/v2 = 7/6
So a+b = 13