Question

Three particles start from origin at the same time, one with velocity $v_1$ along positive x-axis, the second along the positive y-axis with a velocity $v_2$ and the third along the line y = x with such a speed that all the three remain on a straight line. The velocity of the third particle, is

• A. $\sqrt{v_1v_2}$
• B. $\frac{v_1+v_2}{2}$
• C. $\frac{v_1v_2}{\sqrt{v_1^2+v_2^2}}$
• D. $\frac{v_1v_2\sqrt{2}}{v_1+v_2}$

Answer 1

Answer: (D)

(4)

Answer 2

The problem asks for the velocity of a third particle such that it always remains collinear with two other particles starting from the origin.

Let the time elapsed be $t$. All particles start from the origin (0,0).

1. Position of the first particle ($P_1$): It moves along the positive x-axis with velocity $v_1$. Its position at time $t$ is $P_1 = (v_1 t, 0)$.

2. Position of the second particle ($P_2$): It moves along the positive y-axis with velocity $v_2$. Its position at time $t$ is $P_2 = (0, v_2 t)$.

3. Position of the third particle ($P_3$): It moves along the line $y=x$. Let its velocity be $v_3$. Since it moves along $y=x$, its x and y components of velocity must be equal. Let these components be $v_{3x}$ and $v_{3y}$. So, $v_{3x} = v_{3y}$. The magnitude of its velocity is $v_3 = \sqrt{v_{3x}^2 + v_{3y}^2}$. Substituting $v_{3y} = v_{3x}$, we get $v_3 = \sqrt{v_{3x}^2 + v_{3x}^2} = \sqrt{2v_{3x}^2} = v_{3x}\sqrt{2}$. Therefore, $v_{3x} = \frac{v_3}{\sqrt{2}}$. The position of the third particle at time $t$ is $P_3 = (x_3, y_3)$, where $x_3 = v_{3x} t = \frac{v_3 t}{\sqrt{2}}$ and $y_3 = v_{3y} t = \frac{v_3 t}{\sqrt{2}}$. So, $P_3 = \left(\frac{v_3 t}{\sqrt{2}}, \frac{v_3 t}{\sqrt{2}}\right)$.

Condition for collinearity: For the three particles to remain on a straight line, their positions $P_1$, $P_2$, and $P_3$ must be collinear for all $t > 0$. We can use the intercept form of the equation of a straight line. The line passing through $P_1(v_1 t, 0)$ and $P_2(0, v_2 t)$ has x-intercept $v_1 t$ and y-intercept $v_2 t$. The equation of this line is: $\frac{x}{v_1 t} + \frac{y}{v_2 t} = 1$ To simplify, multiply the entire equation by $v_1 v_2 t$: $v_2 x + v_1 y = v_1 v_2 t$ Now, substitute the coordinates of $P_3\left(\frac{v_3 t}{\sqrt{2}}, \frac{v_3 t}{\sqrt{2}}\right)$ into this equation: $v_2 \left(\frac{v_3 t}{\sqrt{2}}\right) + v_1 \left(\frac{v_3 t}{\sqrt{2}}\right) = v_1 v_2 t$ Since this equation must hold for any time $t > 0$, we can divide both sides by $t$: $v_2 \frac{v_3}{\sqrt{2}} + v_1 \frac{v_3}{\sqrt{2}} = v_1 v_2$ Factor out $\frac{v_3}{\sqrt{2}}$ from the left side: $\frac{v_3}{\sqrt{2}} (v_1 + v_2) = v_1 v_2$ Solve for $v_3$: $v_3 = \frac{v_1 v_2 \sqrt{2}}{v_1 + v_2}$

Comparing this result with the given options: (1) $\sqrt{v_1v_2}$ (2) $\frac{v_1+v_2}{2}$ (3) $\frac{v_1v_2}{\sqrt{v_1^2+v_2^2}}$ (4) $\frac{v_1v_2\sqrt{2}}{v_1+v_2}$

The calculated velocity matches option (4).

The final answer is (4)

Explanation of the solution:

1. Determine the position vectors of the first two particles at time t based on their velocities and starting point (origin). $P_1 = (v_1 t, 0)$ and $P_2 = (0, v_2 t)$.
2. Determine the position vector of the third particle at time t. Since it moves along y=x with velocity $v_3$, its velocity components are $v_{3x} = v_{3y} = v_3/\sqrt{2}$. Thus, $P_3 = (v_3 t/\sqrt{2}, v_3 t/\sqrt{2})$.
3. For the three particles to remain collinear, their positions must satisfy the equation of a straight line. Use the intercept form of the line passing through $P_1$ and $P_2$: $\frac{x}{v_1 t} + \frac{y}{v_2 t} = 1$.
4. Substitute the coordinates of $P_3$ into this line equation and solve for $v_3$. This yields $v_3 = \frac{v_1 v_2 \sqrt{2}}{v_1 + v_2}$.