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Prove that in case of oblique elastic collision of two particles of equal mass if one is at rest, the recoiling particles always move off at right angles to each other. |
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Answer» Solution :Method - I: In elastic collision momentum is conserved, So, conservation of momentum along x-axis yields `m u = m v _(1) cos theta _(1) + m v _(2) cos theta_(2) , i.e., u = v _(1) cos theta_(1) + v _(2) cos theta _(2)….(1)` and along y-axis yields, `0 = v _(1) sin thea _(1) - v _(2) sin theta _(2) ....(2)` Squaring and adding Eqns. (1) and (2), we get `u^(2) = v _(1) ^(2) + v _(2) ^(2) + 2 v _(1) v _(2) cos (theta _(1) + theta _(2)).....(3)` As the collision is elastic `1/2 m u ^(2) = 1/2 mv _(1) + 1/2 mv _(2) ^(2)` `i.e., u ^(2) = v _(1) ^(2) + v _(2) ^(2) ....(4)` In the light of Eqn. (4), (3) reduces to `2 v _(1) v _(2) cos (theta _(1) + theta_(2)) = 0` As it is given that `v _(1) ne 0 and v _(2) ne 0` so ` cos (theta _(1) + theta_(2)) =0 implies (theta_(1) + theta_(2)) = 90^(@)`  Method -II: In head on elastic collision between TWO particles, they exchange their VELOCITIES. In thsi case, the component of ball 1 along common normal direction, `v cos theta`  becomes zero after collision, wheile that of 2 becomes v `cos theta.` While the components along common tangent directio of particles remain unchanged. Thus, the components along common tangent and common normal direction of both the balls in TABULAR form are given below.  From the above table and figure, we SEE that both the balls move at right angles after collision with velocities ` v sin theta and v cos theta.` |
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