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(C) Conservation of linear momentum along the $X$-axis:$mu_1 = mv_1 \cos 	heta_1 + mv_2 \cos 	heta_2 \implies u_1 = v_1 \cos 	heta_1 + v_2 \cos 	h

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(C) Conservation of linear momentum along the $X$-axis:$mu_1 = mv_1 \cos 	heta_1 + mv_2 \cos 	heta_2 \implies u_1 = v_1 \cos 	heta_1 + v_2 \cos 	heta_2 \quad \dots(i)$Conservation of linear momentum along the $Y$-axis:$0 = mv_1 \sin 	heta_1 - mv_2 \sin 	heta_2 \implies v_1 \sin 	heta_1 = v_2 \sin 	heta_2 \quad \dots(ii)$Conservation of kinetic energy:$\frac{1}{2}mu_1^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 \implies u_1^2 = v_1^2 + v_2^2 \quad \dots(iii)$Squaring and adding equations $(i)$ and $(ii)$:$u_1^2 = (v_1 \cos 	heta_1 + v_2 \cos 	heta_2)^2 + (v_1 \sin 	heta_1 - v_2 \sin 	heta_2)^2$$u_1^2 = v_1^2 \cos^2 	heta_1 + v_2^2 \cos^2 	heta_2 + 2v_1v_2 \cos 	heta_1 \cos 	heta_2 + v_1^2 \sin^2 	heta_1 + v_2^2 \sin^2 	heta_2 - 2v_1v_2 \sin 	heta_1 \sin 	heta_2$$u_1^2 = v_1^2(\cos^2 	heta_1 + \sin^2 	heta_1) + v_2^2(\cos^2 	heta_2 + \sin^2 	heta_2) + 2v_1v_2(\cos 	heta_1 \cos 	heta_2 - \sin 	heta_1 \sin 	heta_2)$$u_1^2 = v_1^2 + v_2^2 + 2v_1v_2 \cos(	heta_1 + 	heta_2)$Comparing with equation $(iii)$ $(u_1^2 = v_1^2 + v_2^2)$:$v_1^2 + v_2^2 = v_1^2 + v_2^2 + 2v_1v_2 \cos(	heta_1 + 	heta_2)$$2v_1v_2 \cos(	heta_1 + 	heta_2) = 0$Since $v_1, v_2 
eq 0$,$\cos(	heta_1 + 	heta_2) = 0$$	heta_1 + 	heta_2 = 90^o$.

$A$ body of mass $m$ moving with velocity $u_1$ collides elastically with another body of the same mass $m$ at rest. After the collision,they move at an angle of .............. $^o$ to each other.

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