Question

Robert, standing at the rear end of a railroad car of length \(100 . \mathrm{m}\), shoots an arrow toward the front end of the car. He measures the velocity of the arrow as \(0.300 c .\) Jenny, who was standing on the platform, saw all of this as the train passed her with a velocity of \(0.750 c\). Determine the following as observed by Jenny: a) the length of the car b) the velocity of the arrow c) the time taken by arrow to cover the length of the car d) the distance covered by the arrow

Short answer

Answer: As observed by Jenny: a) The length of the car is 66.07 m b()) The velocity of the arrow is 0.857c c) The time taken by the arrow to cover the length of the car is \(1.674 \times 10^{-9}\) s d) The distance covered by the arrow is 431.28 m

Step-by-step solution

Part (a): Finding the length of the car

To find the length of the car as observed by Jenny, we will use the length contraction formula given by: \( L = L_0 \sqrt{1 - (v^2/c^2)} \) Where \(L\) is the length as observed by Jenny, \(L_0\) is the length in the car's rest frame (100 m), and \(v\) is the relative velocity of the train (0.750c). Plugging in the given values, we find the length of the car: \( L = 100 \cdot \sqrt{1 - (0.750c^2/c^2)} = 100 \cdot \sqrt{1 - 0.5625} = 100 \cdot \sqrt{0.4375} = 66.07 \, m\)

Part (b): Finding the velocity of the arrow

To find the velocity of the arrow as observed by Jenny, we will use the velocity addition formula given by: \( u = \frac{u_0 + v}{1 + \frac{u_0 v}{c^2}} \) Where \(u\) is the velocity of the arrow as observed by Jenny, \(u_0\) is the velocity in the car's rest frame (0.300c), and \(v\) is the relative velocity of the train (0.750c). Plugging in the given values, we find the velocity of the arrow: \( u = \frac{0.300c + 0.750c}{1 + \frac{(0.300c)(0.750c)}{c^2}} = \frac{1.050c}{1 + 0.225} = \frac{1.050c}{1.225} = 0.857c \)

Part (c): Finding the time taken by the arrow

To find the time taken by the arrow to cover the length of the car, we will first find the time taken in the car's frame (stationary to Robert) and then use the time dilation formula to find the time in Jenny's frame. In the car's frame, we have: \(t_0 = \frac{L_0}{u_0} = \frac{100 \, m}{0.300c} = \frac{100 \, m}{0.300 \cdot (3 \times 10^8 \, m/s)} = 1.111 \times 10^{-9}\, s\) Now we will use the time dilation formula given by: \( t = \frac{t_0}{\sqrt{1 - (v^2/c^2)}} \) Plugging in the given values, we find the time taken: \( t = \frac{1.111 \times 10^{-9}\, s}{\sqrt{1 - (0.750c^2/c^2)}} = 1.674 \times 10^{-9}\, s \)

Part (d): Finding the distance covered by the arrow

To find the distance covered by the arrow as observed by Jenny, we will use the definition of velocity: \( d = ut \) Plugging in the values we found earlier: \( d = (0.857c)(1.674 \times 10^{-9}\, s) = (0.857 \cdot 3 \times 10^8 \, m/s)(1.674 \times 10^{-9}\, s) = 431.28 \, m \) So, as observed by Jenny: a) The length of the car is 66.07 m b) The velocity of the arrow is 0.857c c) The time taken by the arrow to cover the length of the car is \(1.674 \times 10^{-9}\) s d) The distance covered by the arrow is 431.28 m