JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A body of mass 14 kg initially at rest explodes and breaks into three fragments of masses in the ratio \(2:2:3\). The two pieces of equal masses fly off perpendicular to each other with a speed of \(18~m/s\) each. The velocity of the heavier fragment is ___________ \(m/s\).
- A \(10\sqrt{2}\)
- B \(12\sqrt{2}\)
- C 12
- D \(24\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(12\sqrt{2}\)
Step-by-step Solution
Detailed explanation
Apply C.O.M \(M_{1}V_{1}+M_{2}V_{2}+M_{3}V_{3}=0\) \(2(-18i)+2(-18j)+3V_{3}=0\) \(V_{3}=12i+12j\) \(|V_{3}|=12\sqrt{2}m/s\)
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