**49.** The position vectors of two 1 kg particles, (A) and (B), are given by: \[ \vec{r}_A = \left( \alpha_1 t^2 \hat{i} + \alpha_2 t \hat{j} + \alpha_3 t \hat{k} \right) \, \text{m} \] and \[ \vec{r}_B = \left( \beta_1 t \hat{i} + \beta_2 t^2 \hat{j} + \beta_3 t \hat{k} \right) \, \text{m}, \] respectively; where: \[ \alpha_1 = 1 \, \text{m/s}^2,\quad \alpha_2 = 3n \, \text{m/s},\quad \alpha_3 = 2 \, \text{m/s}, \beta_1 = 2 \, \text{m/s},\quad \beta_2 = -1 \, \text{m/s}^2,\quad \beta_3 = 4p \, \text{m/s} \] \( t \) is time, and \( n \), \( p \) are constants. At \( t = 1 \, \text{s} \), \( |\vec{V}_A| = |\vec{V}_B| \), and the 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 \( \sqrt{L} \, \text{kg·m}^2\text{/s} \). The value of \( L \) is _____.

1. Turning position into velocity

A particle’s velocity is the time-derivative of its position. If $\vec r_A(t)= (\alpha_1 t^2)\hat i + (\alpha_2 t)\hat j + (\alpha_3 t)\hat k$ then $\vec v_A(t)=\frac{d\vec r_A}{dt}=2\alpha_1 t\,\hat i + \alpha_2\,\hat j + \alpha_3\,\hat k.$ Do exactly the same for particle B.

2. Using the equal–speed condition

Equal speeds at $t=1\,\text{s}$ means $|\vec v_A(1)| = |\vec v_B(1)|.$ Because speed is the magnitude of velocity, we square both sides to avoid radicals and get a simple relation between $n$ and $p$.

3. Using the perpendicular condition

Two vectors are orthogonal when their dot product is zero: $\vec v_A(1)\cdot \vec v_B(1)=0.$ Applying this at $t=1\,\text{s}$ gives the second equation in $n$ and $p$.

With two independent equations we can now solve for the two unknowns.

4. Angular momentum about a moving point

The angular momentum of particle A about the instantaneous position of B is $\vec L= (\vec r_A-\vec r_B)\times (m\,\vec v_A).$ Here $m=1\,\text{kg}$, so it is simply the cross product of the displacement vector and A’s velocity.

5. Magnitude

The problem finally wants $|\vec L| = \sqrt{L}\qquad \Longrightarrow \qquad L=|\vec L|^{2}.$ So once the cross product components are known, square them, add them, and that sum is $L$.

What is the question?

Two tiny 1-kg balls (call them A and B) are flying through space. You are told exactly where each ball is at any second t by giving its $x$, $y$, $z$ coordinates. From those, you can find how fast each ball is moving (its velocity), how the two velocities compare, and how much "spin" (angular momentum) ball A has as seen from ball B.

What must be true at $t = 1\,\text{s}$?

1. The speeds of the two balls are the same.
2. The two velocity arrows are at right angles to each other ("orthogonal").

That lets us solve for the two unknown numbers $n$ and $p$ hidden inside the given position formulas.

Finally, we plug those $n$ and $p$ back in, work out the angular-momentum arrow (a cross-product of the relative-position arrow and velocity arrow), find its length, and then square it. That square is called $L$, and the problem asks for that number.

Think of it like two kids on skateboards: they are moving at the same speed but one is going east while the other is going north, and we want to know how much spin one kid has as seen from the other.

1. Velocities at $t=1\,\text{s}$
   From the derivatives,
   $\vec v_A(1)=2\hat i+3n\hat j+2\hat k,$
   $\vec v_B(1)=2\hat i-2\hat j+4p\hat k.$

2. Equal speeds
   |\vec v_A|^2=|\vec v_B|^2 \;\\ \Rightarrow\; 4+9n^2+4 = 4+4+16p^2 \;\\ \Rightarrow\; 9n^2 = 16p^2 \;\Longrightarrow\; \left|\frac{n}{p}\right| = \frac43. \tag{1}

3. Perpendicular velocities
   \vec v_A\cdot\vec v_B = 0 \;\\ \Rightarrow\; (2)(2) + (3n)(-2) + (2)(4p) = 0 \;\\ \Rightarrow\; 4 - 6n + 8p = 0. \tag{2}

4. Solve (1) and (2)
   Choosing $n = -\frac43 p$ (the only choice compatible with (2)) gives
   $6\Bigl(-\tfrac43 p\Bigr) - 8p = 4 \;\Longrightarrow\; -16p = 4 \;\Longrightarrow\; p = -\tfrac14,\qquad n = \tfrac13.$

5. Relative position at $t=1\,\text{s}$
   $\vec r_A(1)=\hat i+1\hat j+2\hat k,\qquad \vec r_B(1)=2\hat i-1\hat j-1\hat k,$
   $\vec r = \vec r_A-\vec r_B = -\hat i + 2\hat j + 3\hat k.$

6. Angular momentum

   \begin{vmatrix} \hat i & \hat j & \hat k\\ -1 & 2 & 3\\ 2 & 1 & 2 \end{vmatrix} = \hat i(4-3) - \hat j(-2-6) + \hat k(-1-4) = \hat i + 8\hat j -5\hat k.$$

7. Magnitude and $L$
   $|\vec L| = \sqrt{1^2 + 8^2 + (-5)^2} = \sqrt{90} = 3\sqrt{10},\qquad L = (|\vec L|)^2 = 90.$

[\boxed{L = 90}]