A body of mass 10 kg explodes at rest into three fragments with masses in the ratio $1 : 4 : 5$. The

Question

A body of mass 10 kg explodes at rest into three fragments with masses in the ratio $1 : 4 : 5$. The fragments with masses in the ratio $4 : 5$ fly in mutually perpendicular directions with speeds of $15 \mathrm{~m/s}$ and $12 \mathrm{~m/s}$ respectively. The velocity of the lightest fragment in $\mathrm{m/s}$ will be

Professor PC Thomas Classes · Answer

Mass ratio 1:4:5 for 10 kg means fragments are 1 kg, 4 kg, 5 kg. The 4 kg fragment moves at 15 m/s ($p_1 = 60$). The 5 kg fragment moves at 12 m/s ($p_2 = 60$). The resultant momentum of these two is $p_r = \sqrt{60^2 + 60^2} = 60\sqrt{2}$. By conservation of momentum, $p_3 = p_r$. So $1 	imes v_3 = 60\sqrt{2} \approx 60 	imes 1.414 = 84.84 \mathrm{~m/s}$.