Question

Q1.a bomb at rest explodes into three parts of the same mass the momentum of the two points are X i cap and - 2 x j cap the momentum of the third part will have a magnitude of _______?

Answer 1

Step-by-step explanation:

Answer: x√5 Units

Explanation:

Let the momentum of the bomb before being exploded be P and momentum of it's three parts after exploding be P₁ , P₂ , P₃ respectively.

Since the bomb was under rest before exploding therefore,

$\begin{gathered}\vec{P}=0\\\;\\\vec{P_1}=x\hat{i}\\\;\\\vec{P_2}=-2\hat{j}\end{gathered}$

According to law of conversation of linear momentum,

$\begin{gathered}\vec{P_1}+\vec{P_2}+\vec{P_3}=\vec{P}\\\;\\x\hat{i}-2x\hat{j}+\vec{P_3}=0\\\;\\\vec{P_3}=-x\hat{i}+2x\hat{j}\\\;\\\text{Taking modulus of both sides}\\\;\\P_3=\sqrt{(-x)^2+(2x)^2}\\\;\\P_3=\sqrt{x^2+4x^2}\\\;\\P_3=\sqrt{5x^2}\\\;\\P_3=x\sqrt{5}\;\;\;\text{Units}\end{gathered}$

To find direction of P₃:-

$\begin{gathered}\text{Since we have,}\\\;\\\vec{P_3}=-x\hat{i}+2\hat{j}\\\;\\\frac{\vec{P_3}}{\mid P_3\mid}=\frac{-x\hat{i}}{\mid P_3\mid}+\frac{2\hat{j}}{\mid P_3\mid}\\\;\\\hat{P_3}=\frac{-x\hat{i}}{x\sqrt{5}}+\frac{2\hat{j}}{x\sqrt{5}}\\\;\\\hat{P_3}=(-\frac{1}{\sqrt{5}})\hat{i}+(\frac{2}{\sqrt{5}})\hat{j}\end{gathered}$

Answer 2

above answer is correct