Three blocks are placed on smooth horizontal surface and lie on same horizontal straight line.Block 1 and block 3 have mass m each and block 2 has mass M(M greater than m) .Block2 and block3 are initially stationary ,while block1 is initially moving towards block2 with speed v as shown.Assume that all collisions are head on and perfectly elastic .What value of M/m ensures that block1 and block 3 have the same final speed?
Asked by m.nilu
15th September 2018,
7:46 PM
Answered by Expert
Answer:
Initially collision is taking place between block-1 and block-2
let v1 and v2 are velcoties after collison for block-1 and block-2 respectively. Let α = M/m .
Due to momentum conservation, 
Due to energy conservation, 
By solving (1) and (2), we get
Now block-2 collides with block-3.
Let v3 and v4 are respective velocities of block-2 and block-3 after collisions.
By Coservation of Momentum, 
....................(5)
....................(5)By Coservation of energy, 
...........................(6)
...........................(6)using eqns. (5) and (6), we get quadratic eqn. 
...................(7)
...................(7)Two solutions of the above quadratic equation are, v3 = v2 and 
the solution v3 = v2 is unacceptable because it gives v4 = 0. Hence if we consider the other solution, we get 
Now if we equate the magnitude of v4 and v1 , we get 
............................(8)
............................(8)[ it is to be noted if α>1, sign of v1 is negative. hence we compared the magnitude of v1 and v4 in eqn.(8) ]
Solving eqn.(8) for α , we get an acceptable solution, α = 2+√5 or (M/m) = 2+√5
Answered by Expert
20th September 2018,
12:15 PM
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