The position vectors of two 1 kg particles, (A) and (B), are given by and , respectively; , , where t is time, n and p are constants, At t = 1 s, and velocities and of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular mo

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The position vectors of two 1 kg particles, (A) and (B), are given by $\vec{r_{A}} = (α_{1} t^{2} \hat{i} + α_{2} \hat{j} + α_{3} t \hat{k}) m$ and $\vec{r_{B}} = (β_{1} t \hat{i} + β_{2} t^{2} \hat{j} + β_{3} \hat{t} \hat{k}) m$ , respectively; $(α_{1} = 1 m / s^{2}, α_{2} = 3 n m / s, α_{3} = 2 m / s, β_{1} = 2 m / s)$ , $(β_{2} = - 1 m / s^{2}, β_{3} = 4 p m / s)$ , where t is time, n and p are constants, At t = 1 s, $| {\vec{V}}_{A} |$ $=$ $| {\vec{V}}_{B} |$ and velocities ${\vec{V}}_{A}$ and ${\vec{V}}_{B}$ of the particles are orthogonal to each other. At t = 1 s, the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is $\sqrt{L} {kgm}^{2} s^{- 1}$ . The value of L is _________. [2025]

Ans.
(90)

${\vec{r}}_{A} = (α_{1} t^{2} \hat{i} + α_{2} \hat{j} + α_{3} t \hat{k}) m$

$\Rightarrow \frac{d {\vec{r}}_{A}}{d t} = {\vec{V}}_{A} = (2 t \hat{i} + 3 n \hat{j} + 2 \hat{k})$

${\vec{r}}_{B} = (β_{1} t \hat{i} + β_{2} t^{2} \hat{j} + β_{3} \hat{t} \hat{k})$

$\Rightarrow \frac{d {\vec{r}}_{B}}{d t} = {\vec{V}}_{B} = (2 \hat{i} - 2 \hat{j} + 4 p \hat{k})$

At t = 1 s, ${\vec{V}}_{A} \cdot {\vec{V}}_{B} = 0$

4 – 6n + 8p = 0

$\Rightarrow 3 n = 2 + 4 p$ ... (i)

At t = 1 s, $| {\vec{V}}_{A} |$ $=$ $| {\vec{V}}_{B} |$

$4 + 9 n^{2} + 4 = 4 + 4 + 16 p^{2}$

From (i) and (ii), $p = \frac{- 1}{4} \Rightarrow n = \frac{1}{3}$

The angular momentum of particle (A) with respect to the position of particle (B)

$| \vec{L} |$ $= m_{A} ({\vec{r}}_{A / B} \times {\vec{V}}_{A})$

${\vec{r}}_{A / B} = (α_{1} - β_{1}) \hat{i} + (α_{2} - β_{2}) \hat{j} + (α_{3} - β_{3}) \hat{k}$

$= (1 - 2) \hat{i} + (1 + 1) \hat{j} + 3 \hat{k}$

$\vec{L} =$ $| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & 3 \\ 2 & 1 & 2 \end{matrix} |$ $= \hat{i} + 8 \hat{j} - 5 \hat{k}$

$| \vec{L} |$ $= \sqrt{1 + 64 + 25} = \sqrt{90}$

Want Us to Email you About Special Offers & Updates?