(A) PROVE THAT IN CASE OF OBLIQUE ELASTIC COLLISION OF TWO PARTICLES OF SAME MASS IF ONE IS AT REST, - Physics - Work Energy And Power

Question

(A) PROVE THAT IN CASE OF OBLIQUE ELASTIC COLLISION OF TWO
PARTICLES OF SAME MASS IF ONE IS AT REST, AFTER COLLISION BOTH
PARTICLES MOVES AT RIGHT ANGLE TO EACH OTHER

Ruhhi Ralhan · Accepted Answer

Dear Student, Consider two particles A and B of equal masses 'm' Particle B collide with A obliquely by making an angle 'Î¸' with X-axis After collision they move by making an angle Î¦ and Î± with x-axis Refer to the following figure,

Using energy conservation: $\frac{1}{2} m (0)^{2} + \frac{1}{2} m u^{2} = \frac{1}{2} {m v}_{1}^{2} + \frac{1}{2} m v_{2}^{2} u^{2} = v_{1}^{2} + v_{2}^{2}$

Conservation of momentum: $X-axis: m (0) + m u \cos θ = m v_{1} \cos α + m v_{2} \cos Φ u \cos θ = v_{1} \cos Φ + v_{2} \cos α (1) Y-axis: m (0) + m u \sin θ = - m v_{1} \sin α + m v_{2} \sin Φ u \sin θ = v_{2} \sin Φ - v_{1} \sin α (2) Now, squaring and adding equations (1) and (2), {u^{2} = (v_{1} \cos Φ + v_{2} \cos α)}^{2} + (v_{2} \sin Φ - v_{1} \sin α)^{2} u^{2} = v_{1}^{2} + v_{2}^{2} + 2 v_{1} v_{2} (\cos α \cos Φ - \sin α \sin Φ) Now, from kinetic energy conservation, u^{2} = v_{1}^{2} + v_{2}^{2} 2 v_{1} v_{2} (\cos α \cos Φ - \sin α \sin Φ) = 0 Using cosine sum formula: \cos A \cos B - \sin A \sin B = \cos (A - B) T h e r e f o r e, \cos (α + Φ) = 0 o r, α + Φ = 90^{0}$ So we have proved after collision particles get apart at perpendicular to each other

(A) PROVE THAT IN CASE OF OBLIQUE ELASTIC COLLISION OF TWO PARTICLES OF SAME MASS IF ONE IS AT REST, AFTER COLLISION BOTH PARTICLES MOVES AT RIGHT ANGLE TO EACH OTHER.