Question

The position vectors of two 1 kg particles, (A) and (B), are given by \[ \vec{r}_A = (\alpha_1 t \hat{i} + \alpha_2 t^2 \hat{j} + \alpha_3 t^3 \hat{k}) \, \text{m} \] and \[ \vec{r}_B = (\beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t^3 \hat{k}) \, \text{m}, \text{ respectively; } \] \[ (\alpha_1 = 1 \, \text{m/s}, \, \alpha_2 = 3 \, \text{m/s}^2, \, \alpha_3 = 2 \, \text{m/s}^3, \, \beta_1 = 2 \, \text{m/s}, \, \beta_2 = -1 \, \text{m/s}^2, \, \beta_3 = 4 \, \text{m/s}^3), \] where \( t \) is time, and \( n \) and \( p \) are constants. At \( t = 1 \, \text{s}, \, |\vec{V}_A| = |\vec{V}_B| \) and velocities \( \vec{V}_A \) and \( \vec{V}_B \) are orthogonal to each other. At \( t = 1 \, \text{s} \), the magnitude of angular momentum of particle (A) with respect to the position of particle (B) is \( \sqrt{L} \, \text{kgm}^2\text{s}^{-1} \). The value of \( L \) is ______.}

Hint:

For angular momentum, use the formula \( \vec{L} = m \, \vec{r} \times \vec{v} \), where \( \vec{r} \) is the position vector and \( \vec{v} \) is the velocity.

Answer 1

Correct Answer: 90

The velocities of particles \( A \) and \( B \) are: \[ \vec{V}_A = 2t \hat{i} + 6nt \hat{j} + 8pt^2 \hat{k} \] \[ \vec{V}_B = 2t \hat{i} - 2t \hat{j} + 4pt^2 \hat{k} \] Since \( \vec{V}_A \cdot \vec{V}_B = 0 \), we get: \[ 4 - 6n + 8p = 0 \] \[ 2 - 3n + 4p = 0 \quad \Rightarrow \quad 3n = 2 + 4p \] Solving this, we get \( p = \frac{-1}{4} \), and \( n = \frac{1}{3} \). 

Thus, the angular momentum \( L \) is: \[ \vec{L} = m \left( \vec{r}_{A/B} \times \vec{V}_A \right) \] 

Using the cross product, we get: $\vec{r}_{A/B} = (\alpha_1 - \beta_1) \hat{i} + (\alpha_2 - \beta_2) \hat{j} + (\alpha_3 - \beta_3) \hat{k}$ $= (1-2) \hat{i} + (1+1) \hat{j} + 3 \hat{k}$ $= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} -1 & 2 & 32 & 1 & 2 \end{vmatrix}$ $= \hat{i}(4-3) - \hat{j}(-2-6) + \hat{k}(-1-4)$ $= \hat{i} + 8\hat{j} - 5\hat{k}$ $= \sqrt{1^2 + 8^2 + (-5)^2}$ 

Thus: \[ \sqrt{1 + 64 + 25} = \sqrt{90} \]