NEET · Physics · STD 11 - 4.1 newtons laws of motion
A shell of mass \(m\) is at rest initially. It explodes into three fragments having mass in the ratio \(2: 2: 1\). If the fragments having equal mass fly off along mutually perpendicular directions with speed \(v\), the speed of the third (lighter) fragment is :
- A \(\sqrt{2} v\)
- B \(2 \sqrt{2} v\)
- C \(3 \sqrt{2} v\)
- D \(v\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2} v\)
Step-by-step Solution
Detailed explanation
By conservation of momentum : \(m(0)=\frac{2 m}{5}(-v \hat{i})+\frac{2 m}{5}(-v \hat{j})+\frac{m}{5} \vec{v}^{\prime}\)
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