A body of mass 8 kg explodes at rest into three fragments with masses in the ratio $1: 3: 4$. The fr

Question

A body of mass 8 kg explodes at rest into three fragments with masses in the ratio $1: 3: 4$. The fragments with masses in the ratio $3: 4$ fly in mutually perpendicular directions with speeds of $21 \mathrm{~m} / \mathrm{s}$. The velocity of heaviest fragment in $\mathrm{m} / \mathrm{s}$ will be\

Professor PC Thomas Classes · Answer

(a) : Since 8 kg body explodes into three fragments with masses in the ratio $1: 3: 4$ thus, masses of fragments will be $1 \mathrm{~kg}, 3 \mathrm{~kg}$ and 4 kg respectively. The magnitude of resultant momentum of two fragments, moving with velocity $21 \mathrm{~m} / \mathrm{s}$, in perpendicular directions is $$ \sqrt{\left(m_{1} v_{1}ight)^{2}+\left(m_{2} v_{2}ight)^{2}}=\sqrt{(3.21)^{2}+(4.21)^{2}}=21*5\mathrm{~kg} \mathrm{~m} / \mathrm{s} $$ According to law of conservation of linear momentum\ $m_{3} v_{3}=105$