The position vectors of two 1 kg particles, (A) and (B), are given by $\vec r_A=(\alpha_1t^2\hat i+\alpha_2t\hat j+\alpha_3t\hat k)$ m and $\vec r_B=(\beta_1t\hat i+\beta_2t^2\hat j+\beta_3t\hat k)$ m, respectively; ($\alpha_1=1$ m/s², $\alpha_2=3n$ m/s, $\alpha_3=2$ m/s, $\beta_1=2$ m/s, $\beta_2=-1$ m/s², $\beta_3=4p$ 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 $\sqrt L$ kg m² s⁻¹. The value of L is _______.

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] At $t=1$: $\vec V_A=(2,3n,2)$ and $\vec V_B=(2,-2,4p)$. $|V_A|=|V_B|\Rightarrow9n^2=16p^2$ and $V_A\cdot V_B=0\Rightarrow3n=2+4p$. These give $3n=-4p$, so $p=-\frac14$ and $n=\frac13$. Then $\vec V_A=(2,1,2)$ and $\vec r_A-\vec r_B=(1,1,2)-(2,-1,-1)=(-1,2,3)$. $\vec L=(\vec r_A-\vec r_B)	imes m\vec V_A=(1,8,-5)$ and $|\vec L|^2=90$, so $L=90$. Answer: 90.

Explanation

Reviewed by Dr. Abubacker

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