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In example 31 consider and elastic collision between a neutron and a light uucle like carbon and colculate fractional KE lost. |
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Answer» Solution :`f_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2)` using the result of solved EXAMPLE 31 `m_(2)=12m_(1)` `f_(1)=((m_(1)-m_(2))/(m_(1)+m_(2)))^(2),` Using the result of solved example 18. `f=((m_(1)-12m_(2))/(m_(1)+12m_(1)))^(2),` Using `m_(2)=12m_(1)` `=((-11)/(13))^(2)` `=121/169` Expressed as percent it is `(12100)/(169)=71.6%` and `f_(2)=28.4%` In PARTICLE, however this NUMBER is smaller as head-on COLLISION are none. |
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