a train travels in the +x direction with a speed of =0.80 with respect to the gr
ID: 1999870 • Letter: A
Question
a train travels in the +x direction with a speed of =0.80 with respect to the ground. At a certain time, two balls are ejected, one traveling in the +x direction with x-velocity of +0.60 with respect to the train and the other traveling in the -x direction with x-velocity of -0.40 with respect to the train. (a) What are the x-velocities of the balls with respect to the ground? (b) what is the x-velocity of the first ball with respect to the second? [hint: the frames you will choose to be the Home Frame and the Other Frame for part (a) will not be the same as your choices for part (b).]
Explanation / Answer
What you say about finding the components and then taking the square root of the sum of the squares is correct. The component of speed in the direction of motion is given as a beta of 0.8. What we need is the component of speed in the y-direction.
The distances as measured relative to the ground or the train will be the same in a direction perpendicular to the direction of motion. Call these D, relative to the train, and d relative to the ground. We have D = d.
Using the same upper and lower case conventions for time and speed, we can write:
Relative to train: V = D/T = d/T
Relative to ground: v = d/t
The time as measured on the train will be less than the time as measured relative to ground.
Using b for beta, the relationship between T and t is given by T = t*sqrt(1 - b^2).
V = d/T = d/[t*sqrt(1 - b^2)] = v/[t*sqrt(1 - b^2)]
v = V*sqrt(1 - b^2)
a) Now for the numbers. V = 0.6
sqrt(1 - b^2) = sqrt(1 - 0.8^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6
v = 0.6*0.6 = 0.36
Sum of the squares.
Speed = sqrt(0.8^2 + 0.36^2) = 0.8732
and V = 0.4
sqrt(1 - b^2) = sqrt(1 - 0.8^2) = sqrt(1 - 0.64) = sqrt(0.36) = 0.6
v = 0.4*0.6 = 0.24
Sum of the squares.
Speed = sqrt(0.8^2 + 0.24^2) = 0.8352
b) V1/V2 =0.8732/0.8352 =1.045
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