Two particles A and B move with constant velocities $\vec{v}_{1}$ and $\vec{v}_{2}$. At the initial moment their position vectors $\vec{r}_{1}$ and $\vec{r}_{2 j}$ respectively. The condition for particles $A$ and $B$ for their collision is: (1) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$ (2) $\frac{\vec{r}_{1}-\vec{r}_{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\left|\vec{v}_{2}-\vec{v}_{1}\right|}$ (3) $\overrightarrow{\mathrm{r}}_{1} \cdot \overrightarrow{\mathrm{v}}_{1}=\overrightarrow{\mathrm{r}}_{2} \cdot \overrightarrow{\mathrm{v}}_{2}$ (4) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$ - Noon Academy

Two particles A and B move with constant velocities $\vec{v}_{1}$ and $\vec{v}_{2}$ . At the initial moment their position vectors $\vec{r}_{1}$ and $\vec{r}_{2 j}$ respectively. The condition for particles $A$ and $B$ for their collision is: (1) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$ (2) $\frac{\vec{r}_{1}-\vec{r}_{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\left|\vec{v}_{2}-\vec{v}_{1}\right|}$ (3) $\overrightarrow{\mathrm{r}}_{1} \cdot \overrightarrow{\mathrm{v}}_{1}=\overrightarrow{\mathrm{r}}_{2} \cdot \overrightarrow{\mathrm{v}}_{2}$ (4) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$

Physics | Physics

Two particles A and B move with constant velocities $\vec{v}_{1}$ and $\vec{v}_{2}$ . At the initial moment their position vectors $\vec{r}_{1}$ and $\vec{r}_{2 j}$ respectively. The condition for particles $A$ and $B$ for their collision is: (1) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$ (2) $\frac{\vec{r}_{1}-\vec{r}_{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\left|\vec{v}_{2}-\vec{v}_{1}\right|}$ (3) $\overrightarrow{\mathrm{r}}_{1} \cdot \overrightarrow{\mathrm{v}}_{1}=\overrightarrow{\mathrm{r}}_{2} \cdot \overrightarrow{\mathrm{v}}_{2}$ (4) $\vec{r}_{1} \times \vec{v}_{1}=\vec{r}_{2} \times \vec{v}_{2}$

The correct Solution is (2)
For two particles to collide, the direction of the relative velocity of one with respect to other should be directed towards the relative position of the other particle
i.e. $\frac{\vec{r}_{1}-\vec{r}_{2}}{\left|\vec{r}_{1}-\hat{r}_{2}\right|} \rightarrow$ direction of relative position of 1 w.r.t. 2 .
$\& \frac{\vec{v}_{2}-\vec{v}_{1}}{\left|\vec{v}_{2}\right|-\bar{v}_{1}} \rightarrow$ direction of velocity of 2 w.r.t. 1
So for collision of $A \& B$ ]
$\frac{\vec{r}_{1}-\vec{r}_{2}}{\left|\vec{r}_{1}-\vec{r}_{2}\right|}=\frac{\vec{v}_{2}-\vec{v}_{1}}{\left|\vec{v}_{2}-\vec{v}_{1}\right|}$

i

More Material

The refractive index of glass is $1.5$ . What is the speed of light in glass? Speed of light in vacuum is $3.0 \times 10^{\mathrm{A}} \mathrm{ms}^{-1}$ .

Consider interference between two sources of intensities I and 4I. What will be the intensity at points where phase differences are:

Write two points of difference between interference and diffraction.

The frequency at which the impedance of the circuit boorens mavimum is
a $\frac{1}{2 \pi} \sqrt{\frac{1}{L C}+\frac{R^{2}}{L^{2}}}$
b $\frac{1}{2 \pi} \frac{1}{\sqrt{1 C}}$
c $\frac{1}{2 \mathrm{z}} \sqrt{\frac{1}{\mathrm{LC}}-\frac{\mathrm{R}^{2}}{\mathrm{~L}^{2}}}$
D $\frac{1}{2 \mathrm{e}} \frac{\mathrm{R}}{\mathrm{L}}$

State whether the following is true or false: The magnetic field inside a long circular coil carrying current will be parallel straight lines.

A galvanometer with resistance $100 \Omega$ gives full scale deflection with a current of $10 \mathrm{~mA}$ . The value of shunt, in order to convert it into an ammeter of 10 ampere range, will be
A $-10 \Omega$
B $1 \Omega$
C $0.1 \Omega$
D $0.01 \Omega$

Derive an expression for electric potential at a point due to an electric dipole. Discuss the special cases.

Derive an expression for the potential at a point due to a short dipole. Hence show what will be the potential at an axial and an equatorial point :

Find out the wavelength at which a black body at temperature $700 \mathrm{~K}$ emits maximum energy

Name the type of lens which is used to correct? i) Myopia ii) Hypermetropia.

How does concave lens correct myopia?
A Increases the curvature of the eye lens
B Forms the image of the object at the retina
C Decreases the elongation of the eyeball
D All of the above Correct option is (B) Forms the image of the object at the retina

Explain end-on position and broad side-on position of a bar magnet.

← Previous Next →