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(A) At the highest point,the velocity of the projectile just before the explosion is $v_x = u \cos 53^{\circ} = 50 	imes 0.6 = 30 \, m/s$.Let the mas

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(A) At the highest point,the velocity of the projectile just before the explosion is $v_x = u \cos 53^{\circ} = 50 	imes 0.6 = 30 \, m/s$.Let the mass of the projectile be $2m$. At the highest point,the momentum is $P_i = (2m)v_x = 60m$.After the explosion,one part of mass $m$ comes to rest $(v_1 = 0)$,and the other part of mass $m$ moves with velocity $v_2$.By conservation of linear momentum: $P_i = P_f \implies 60m = m(0) + m(v_2) \implies v_2 = 60 \, m/s$.The radius of curvature is given by $R = \frac{v^2}{a_n}$,where $a_n$ is the normal acceleration.At the highest point,the only acceleration acting is gravity $(g = 10 \, m/s^2)$,which is perpendicular to the velocity.Radius of curvature just before explosion: $R_1 = \frac{v_x^2}{g} = \frac{30^2}{10} = 90 \, m$.Radius of curvature just after explosion for the moving particle: $R_2 = \frac{v_2^2}{g} = \frac{60^2}{10} = 360 \, m$.The ratio is $\frac{R_1}{R_2} = \frac{90}{360} = \frac{1}{4}$.

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