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Solution :

Mass of neutron`~~` mass of proton `=m` <br> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RES_MOD_PHY_XII_C01_S01_032_S01.png" width="80%"> <br> From conservation of momentum in y-direction <br> `sqrt(2mK_(1)) sin 45^(@)=sqrt(2mK_(2)) sin theta`.....(i) <br> In x-direction `sqrt(2mK_(0))-sqrt(2mK_(1))cos 45^(@)=sqrt(2mK_(2))cos theta`.....(ii) <br> Squaring and adding equation (i) and (ii), we have <br> `K_(2)=K_(1)+K_(0)-sqrt(2K_(0)K_(1))`......(iii) <br> From conservation of energy <br> `K_(2)=K_(0)-K_(1)`.......(iv) <br> Solving equations (iii) and (iv), we get <br> `K_(1)=(K_(0))/(2)` <br> i.e., after each collision energy remains half. Therefore, after `n` collisions, <br> `K_(n)=K_(0)((1)/(2))^(n)` <br> `:. 0.23-(4.6xx10^(6))((1)/(2))^(n) 2^(n)=(4.6xx10^(6))/(0.23)` <br> Taking log and solving, we get <br> `n ~~ 24`