Question

relative to the ground, an object is moving rightward with a speed ( v_0 ). three observers each have a different constant velocity relative to the ground as follows: observer 1 is at rest, observer 2 is moving rightward at ( v_0 ), and observer 3 is moving downward at ( v_0 ). which of the observers, if any, will measure the kinetic energy of the object to be the greatest?
a observer 1
b observer 2
c observer 3
d all three observers measure the kinetic energy of the object to be the same.

Explanation:

Step1: Recall the formula for kinetic energy

The formula for kinetic energy is \(K = \frac{1}{2}mv^{2}\), where \(m\) is the mass of the object and \(v\) is the speed of the object relative to the observer.

Step2: Analyze the speed of the object relative to each observer

• Observer 1: The object is moving rightward with speed \(v_{0}\) relative to the ground (and since Observer 1 is at rest relative to the ground), the speed of the object relative to Observer 1 is \(v_{1}=v_{0}\).
• Observer 2: Observer 2 is moving rightward at \(v_{0}\). The relative speed of the object with respect to Observer 2 is \(v_{2}=v_{0}-v_{0} = 0\).
• Observer 3: The object has a right - ward component of speed \(v_{0}\) and Observer 3 is moving downward at \(v_{0}\). Using the Pythagorean theorem for relative velocity (since the motions are perpendicular), the relative speed of the object with respect to Observer 3 is \(v_{3}=\sqrt{v_{0}^{2}+v_{0}^{2}}=\sqrt{2}v_{0}\).

Step3: Compare the kinetic energies

Since \(K=\frac{1}{2}mv^{2}\), and \(m\) is constant:

• \(K_{1}=\frac{1}{2}mv_{0}^{2}\)
• \(K_{2}=\frac{1}{2}m(0)^{2}=0\)
• \(K_{3}=\frac{1}{2}m(\sqrt{2}v_{0})^{2}=mv_{0}^{2}\)

Since \(K_{3}>K_{1}>K_{2}\)

Answer:

C. Observer 3