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`1``sqrt(2)``1/sqrt(2)``2`

Answer :

CSolution :

Just before colisions, velocilty of `m_(1)` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_285_S01.png" width="80%"> <br> `u_(1)=sqrt(2g2a)=2sqrt(ga)` <br> just after collision `m_(1)` is brought to rest and let velocity of `m_(2)` be `v_(2)`. <br>From conservation of linear momentum, `m_(2)v_(2)=m_(1)u_(1)`...........i <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/BMS_VOL2_C01_E01_285_S02.png" width="80%"> <br> Now `v_(2)=sqrt(2ga)` so that `m_(2)` can rise up to point `m`. Putting the value of `u_(1)` and `v_(2)` in eqn i we get <br> `m_(2)sqrt(2ga)=m_(1)2sqrt(ga)impliesm_(1)/m_(2)=1/sqrt(2)`