The friction coefficient between the horizontal surface and each of the blocks shown in figure (9-E20) is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest.
Take g = 10 m/s2.



Given that the velocity of m=2kg block on reaching the 4kg block before collision is 1 m/s,


The velocity of M=4kg block is u2= o.


The friction coefficient between the horizontal surface and each of the blocks is .


Distance between the blocks is s=16 m.


From work energy principle






Since it is a perfectly elastic collision


Let v1 and v2 are the velocities after collision of block 2kg and 4kg respectively.


By law of conservation of momentum




______ (1)


Again,



_______ (2)


Subtracting equation (2) from (1)




Therefore,



Putting work energy principle for 2kg block when comes to rest





Whereare distances travelled by respective blocks.


Putting work energy principle for 4kg block when comes to rest





So, total distance between 2kg and 4kg block is s




Thus, the separation between the two blocks when they come to rest is 5 cm.


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