Question

A point object approaches a stationary plane mirror with speed 10 m/s. As seen by a stationary observer, the image appears to recede from the mirror with speed 6 m/s. What are the direction and magnitude of the image's velocity (as observed by the same stationary observer)?

Answer 1

10 m/s, direction is 6 m/s away from the mirror and 8 m/s parallel to the mirror

Answer 2

Let the velocity of the object be $\vec{v}_O$, the velocity of the image be $\vec{v}_I$, and the velocity of the mirror be $\vec{v}_M$. All velocities are measured with respect to the stationary observer.

The relationship between these velocities for a plane mirror is given by $\vec{v}_I = 2\vec{v}_M - \vec{v}_O$.

The mirror is stationary, so $\vec{v}_M = \vec{0}$. The equation simplifies to $\vec{v}_I = -\vec{v}_O$. This means the velocity of the image is equal in magnitude and opposite in direction to the velocity of the object, as observed by the stationary observer. Therefore, the speed of the image $|\vec{v}_I|$ must be equal to the speed of the object $|\vec{v}_O|$.

We are given that the object approaches the stationary plane mirror with speed 10 m/s. This is the magnitude of the object's velocity as seen by the stationary observer, so $|\vec{v}_O| = 10$ m/s. Since $|\vec{v}_I| = |\vec{v}_O|$, the magnitude of the image's velocity is also 10 m/s.

The problem also states that the image appears to recede from the mirror with speed 6 m/s, as seen by the stationary observer. This refers to the component of the image's velocity perpendicular to the mirror. Let's set up a coordinate system where the x-axis is perpendicular to the mirror and points away from the mirror. Let the mirror be in the y-z plane. The velocity of the image can be decomposed into components perpendicular and parallel to the mirror: $\vec{v}_I = \vec{v}_{I\perp} + \vec{v}_{I\parallel}$. The statement "image appears to recede from the mirror with speed 6 m/s" means the magnitude of the perpendicular component of the image's velocity is 6 m/s, and its direction is away from the mirror. So, $|v_{I\perp}| = 6$ m/s, and since it is receding (moving away from the mirror), $v_{I\perp} = 6$ m/s (in the positive x-direction).

The velocity of the object can also be decomposed: $\vec{v}_O = \vec{v}_{O\perp} + \vec{v}_{O\parallel}$. The relationship $\vec{v}_I = -\vec{v}_O$ implies $\vec{v}_{I\perp} + \vec{v}_{I\parallel} = -(\vec{v}_{O\perp} + \vec{v}_{O\parallel})$. This gives $v_{I\perp} = -v_{O\perp}$ and $\vec{v}_{I\parallel} = -\vec{v}_{O\parallel}$. Using $v_{I\perp} = 6$ m/s, we get $6 = -v_{O\perp}$, so $v_{O\perp} = -6$ m/s. This means the perpendicular component of the object's velocity is 6 m/s towards the mirror, which is consistent with the object "approaching" the mirror.

The total speed of the object is $|\vec{v}_O| = \sqrt{v_{O\perp}^2 + |\vec{v}_{O\parallel}|^2} = 10$ m/s. Substituting $v_{O\perp} = -6$ m/s, we get $\sqrt{(-6)^2 + |\vec{v}_{O\parallel}|^2} = 10$. $36 + |\vec{v}_{O\parallel}|^2 = 100$. $|\vec{v}_{O\parallel}|^2 = 100 - 36 = 64$. $|\vec{v}_{O\parallel}| = 8$ m/s.

Now consider the image velocity $\vec{v}_I = \vec{v}_{I\perp} + \vec{v}_{I\parallel}$. We know $v_{I\perp} = 6$ m/s (away from the mirror). We also know $\vec{v}_{I\parallel} = -\vec{v}_{O\parallel}$. The magnitude is $|\vec{v}_{I\parallel}| = |-\vec{v}_{O\parallel}| = |\vec{v}_{O\parallel}| = 8$ m/s. The direction of the parallel component of the image velocity is opposite to the direction of the parallel component of the object velocity.

The magnitude of the image velocity is $|\vec{v}_I| = \sqrt{v_{I\perp}^2 + |\vec{v}_{I\parallel}|^2} = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ m/s.

The direction of the image's velocity is given by its components: 6 m/s perpendicular to the mirror (away from the mirror) and 8 m/s parallel to the mirror (in the direction opposite to the parallel component of the object's velocity).