49. The position vectors of two 1 kg particles, A and B, are given by: rA = (alpha1 t^2 i + alpha2 t j + alpha3 t k) m and rB = (beta1 t i + beta2 t^2 j + beta3 t k) m, respectively. Where: alpha1 = 1 m/s^2, alpha2 = 3n m/s, alpha3 = 2 m/s; beta1 = 2 m/s, beta2 = -1 m/s^2, beta3 = 4p m/s. t is time, and n, p are constants. At t = 1 s, |VA| = |VB|, and the velocities VA and VB 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) kg·m^2/s. The value of L is _____.

1. Position & Velocity Vectors

A particle’s position in 3-D space is described by a vector

The velocity vector is its time derivative

2. Given data

For particle A (mass 1 kg)

For particle B (mass 1 kg)

Here $n$ and $p$ are unknown constants we must determine.

3. Velocities

Differentiate w.r.t. $t$:

At $t=1\,\text{s}$ we have

4. Conditions at $t=1\,$s

1. Equal speeds: $|\vec v_A|=|\vec v_B|$
2. Orthogonality: $\vec v_A\cdot\vec v_B=0$

These give two equations to find $n$ and $p$.

5. Angular Momentum of A about B

The relative position at any instant is

Linear momentum of A (mass = 1 kg) is simply $\vec p_A=\vec v_A$.

Angular momentum of A about B is

Its magnitude is $|\vec L|$. The problem says this magnitude equals $\sqrt{L}$, so we will eventually square the magnitude to get $L$.

6. Logical flow a student should follow

1. Write velocity vectors at $t=1\,$s.
2. Apply equal-speed condition to relate $n$ and $p$.
3. Apply dot-product-zero condition to get a second relation.
4. Solve the simultaneous equations for $n$ and $p$.
5. Compute positions $\vec r_A$ and $\vec r_B$ at $t=1\,$s.
6. Form the relative vector $\vec r_{AB}$.
7. Evaluate the cross product $\vec r_{AB}\times\vec v_A$.
8. Take its magnitude; square it to match the required form $\sqrt{L}$.

Grasping dot products (orthogonality) and cross products (perpendicular area leading to angular momentum) is the chief conceptual hurdle.

Imagine two tiny balls moving in space

1. Where are they?
   Each ball has a little arrow (called a position vector) telling us exactly where it is at any time.
2. How fast and which way?
   If we look at how that arrow changes, we get another arrow called velocity.
3. What do we know at 1 second?
   • Both balls are equally fast.
   • Their velocity arrows are at right angles (they make an "L").
4. Spinning feeling (angular momentum)
   If you stand on one ball and look at how the other one is moving around you, the whirling strength you feel is called angular momentum.
5. What do we need?
   We must find a number $L$ so that the whirling strength is $\sqrt{L}$.

To do that we plug in the facts (equal speed, right-angle velocities) to discover the mystery constants $n$ and $p$, then use the cross-product rule for angular momentum. Neat algebra and vector tricks give the final answer.

Step 1: Write velocities at $t=1\,$s

Step 2: Equal speeds

Step 3: Orthogonality

Step 4: Solve for $n$ and $p$

Equate the two expressions for $p$:

1. $\dfrac{3n-2}{4}=\dfrac{3}{4}n$ gives $-2=0$ (impossible).

2. $\dfrac{3n-2}{4}=-\dfrac{3}{4}n$ gives

   $3n-2=-3n\;\Rightarrow\;6n=2\;\Rightarrow\;\boxed{n=\dfrac{1}{3}}$

   $p=-\frac{3}{4}\left(\frac{1}{3}\right)=\boxed{-\dfrac{1}{4}}$

Step 5: Positions at $t=1\,$s

Step 6: Relative position

Step 7: Cross product

Using $\vec p_A=\vec v_A(1)=2\,\hat i+1\,\hat j+2\,\hat k$ (mass = 1 kg):

[ \vec L=\begin{vmatrix} \hat i & \hat j & \hat k\ -1 & 2 & 3\ 2 & 1 & 2 \end{vmatrix}=1,\hat i+8,\hat j-5,\hat k ]

Step 8: Magnitude and $L$

Problem states $|\vec L|=\sqrt{L}$, hence

[ \boxed{L=90} ]