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The friction coefficient between the hor...

The friction coefficient between the horizontal surface and each of the block shown in the figure is `0.2`. The collision between the blocks is perfectly elastic. Find the separation between them when they come to rest. (Take `g=10m//s^2`).

Text Solution

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Velocity of first block before collision, `v_(1)^(2)=1^(2)-2(2)xx0.16=1-0.64`
`v_(1)=0.6 m//s`
By conservation of momentum, `2xx0.6=2 v_(1)+4 v._(2)`
also `v._(2)-v._(1)=v_(1)` for elastic collision
If gives `v._(2)=0.4 m//s`
`v._(1)=0.2 m//s`
Now distance moved after collision
`s_(1)=((0.4)^(2))/(2xx2)& s_(2)=((0.2)^(2))/(2xx2)`
`therefore s=s_(1)+s_(2)=0.05m = 5 cm`.
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