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Consider a one diamensional elastic collision between a given incoming body Aand B, initially at rest. How would you choose the mass of B in comparision to the mass of A in order that B should recoil with (a) greatest speed (b) greatest momentum and (c ) greatest kinetic energy ? |
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Answer» Solution :As in a collision, momentum is always CONSERVED `m _(A) u = m _(A) v _(A) + m _(B) u _(B) , u = v _(A) + kv _(B)` (with `k = m _(B) // m _(A))…(1)` Now as collision is elastic `1/2 m _(A) u ^(2) = 1/2 m _(A) v _(A) ^(2) + 1/2 m _(B)v _(B) ^(2) (or)u ^(2) = v _(A) ^(2) + kv _(B) ^(2) ""...(2)` Substituging the VALUE of `v _(A)` from Eqn. (1) and (2) `u ^(2) = (u - kv _(B)) ^(2) + kv _(B) ^(2)` which on solving gives `v _(B) = (2u)/(1 +k)""...(3)` (a) So for `u _(B)` to be max, k must be min. `i.e., k = (m _(B))/( m _(A)) to 0 or m _(B) GT gt m _(A)` (b) Now `P _(a) = m _(B) v_(B) ,` But as `k = (m _(B))/( m _(A)) and v_(B) = (2u)/( 1 + k)` (from Eqn. 3) `P_(a) = (2 km _(A) u )/( 1 + k ) = (2m _(A) u )/(1 + ( 1 //k))` So `p _(B)` will be max when (1/k) is min (or) k is max so `k =(m_(B))/( m _(A)) to OO (or) m _(B) gt gt m _(A)` (c ) `K _(B) = 1/2 m _(B) = v _(B) ^(2)` But as `k = (m _(B))/( m _(A)) and v _(B) = ( 2u)/( 1 + k)` (from Wqn. 3) `K _(B) = 1/2 km _(A) =((2u)/( 1 + K )) ^(2) = (2m _(A) u ^(2) k )/((1+ k ) ^(2)) (or ) K _(B) = ( 4 (K _(A) )k )/( (1- K) ^(2) +4k) [as K _(A) = (1//2) m _(A) u ^(2)]` So, KE will be max when `(1-k)^(2)= min =0,i.e., k = (m _(B))/( m _(A)) =1 (or) m _(B) = m _(A)` |
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