A rigid rod of length 1 m is resting at an angle 45° as shown in the figure. The end P is dragged with a velocity of U = 5 m/s to the right. At the instant shown, the magnitude of the velocity V (in m/s) of point Q as it moves along the wall without losing contact is

The correct answer is 5.

Step-by-step Explanation:

1. Understanding the System and Geometry:
Based on the provided diagram, a rigid rod $PQ$ of length $L=1\text{m}$ slides along a vertical wall (y-axis) and a horizontal floor (x-axis).
Let the coordinates of the ends of the rod be:
- Point $P=(x,0)$
- Point $Q=(0,y)$

At the instant shown in the image, the angle between the rod and the horizontal floor is:
$\theta ={45}^{\circ }$
The velocity of point $P$ along the floor (to the right) is:
$U=\frac{dx}{dt}=5\text{m/s}$
The magnitude of the velocity of point $Q$ along the vertical wall is $V$, where:
$V=-\frac{dy}{dt}$
(since $y$ decreases as point $Q$ moves downwards).

2. Establishing the Constraint Equation:
Since the rod is rigid, its length $L$ remains constant. By the Pythagorean theorem:
$x^2+y^2=L^2$

3. Differentiating with respect to time ($t$):
Differentiating both sides of the constraint equation:
$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$
Simplifying by dividing by 2:
$x\frac{dx}{dt}+y\frac{dy}{dt}=0$

4. Calculating the Velocity $V$:
Substitute the values of the rates of change:
$x\left(U\right)+y(-V)=0$
Rearranging the equation to solve for $V$:
$V=U\left(\frac{x}{y}\right)$
From the geometry of the triangle formed by the wall, floor, and rod, we have:
$\cot \left(\theta \right)=\frac{x}{y}$
Since $\theta ={45}^{\circ }$:
$\frac{x}{y}=\cot \left({45}^{\circ }\right)=1$
Thus, the magnitude of the velocity $V$ is:
$V=U\times 1=5\text{m/s}$