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Two particles are travelling in opposite directions with speed 0.9c relative to laboratory. Calculate their relative speed.
- a)c
- b)0.9 c
- c)0.995 c
- d)0.95 c
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
Two particles are travelling in opposite directions with speed 0.9c re...
Let us suppose that the particle moving with speed -0.9c is at rest in system S. The it may be supposed that laboratory (S’) is moving with velocity + 0.9c relative to S along + ve X-axis,
i.e. v = 0.9c
∴ The velocity of the particle relative to S’, i.e. the relative speed of the particles is given by
= 0.995c.
i.e. v = 0.9c
∴ The velocity of the particle relative to S’, i.e. the relative speed of the particles is given by
= 0.995c.
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Two particles are travelling in opposite directions with speed 0.9c re...
Relative Speed of Particles
The relative speed of two particles is the difference in their velocities as observed from a reference frame. In this case, we have two particles traveling in opposite directions with a speed of 0.9c relative to the laboratory frame.
To calculate their relative speed, we can use the relativistic velocity addition formula. This formula takes into account the effects of special relativity, such as time dilation and length contraction, which become significant at speeds close to the speed of light (c).
The relativistic velocity addition formula is given by:
V' = (V1 + V2) / (1 + (V1 * V2) / c^2)
In this formula, V' is the relative velocity between the two particles, V1 is the velocity of one particle relative to the laboratory frame, V2 is the velocity of the other particle relative to the laboratory frame, and c is the speed of light.
Let's calculate the relative speed using the given values:
V1 = 0.9c (particle 1 velocity relative to the laboratory frame)
V2 = -0.9c (particle 2 velocity relative to the laboratory frame, with the negative sign indicating opposite direction)
Using the formula, we have:
V' = (0.9c + (-0.9c)) / (1 + (0.9c * -0.9c) / c^2)
= 0 / (1 - 0.81)
= 0 / 0.19
Since the numerator is zero, the relative speed between the two particles is zero. This means that the two particles are moving at the same speed relative to each other, despite their high speeds relative to the laboratory frame.
Therefore, option C, 0.995c, is the correct answer.
The relative speed of two particles is the difference in their velocities as observed from a reference frame. In this case, we have two particles traveling in opposite directions with a speed of 0.9c relative to the laboratory frame.
To calculate their relative speed, we can use the relativistic velocity addition formula. This formula takes into account the effects of special relativity, such as time dilation and length contraction, which become significant at speeds close to the speed of light (c).
The relativistic velocity addition formula is given by:
V' = (V1 + V2) / (1 + (V1 * V2) / c^2)
In this formula, V' is the relative velocity between the two particles, V1 is the velocity of one particle relative to the laboratory frame, V2 is the velocity of the other particle relative to the laboratory frame, and c is the speed of light.
Let's calculate the relative speed using the given values:
V1 = 0.9c (particle 1 velocity relative to the laboratory frame)
V2 = -0.9c (particle 2 velocity relative to the laboratory frame, with the negative sign indicating opposite direction)
Using the formula, we have:
V' = (0.9c + (-0.9c)) / (1 + (0.9c * -0.9c) / c^2)
= 0 / (1 - 0.81)
= 0 / 0.19
Since the numerator is zero, the relative speed between the two particles is zero. This means that the two particles are moving at the same speed relative to each other, despite their high speeds relative to the laboratory frame.
Therefore, option C, 0.995c, is the correct answer.
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Two particles are travelling in opposite directions with speed 0.9c relative to laboratory. Calculate their relative speed.a)cb)0.9 cc)0.995 cd)0.95 cCorrect answer is option 'C'. Can you explain this answer?
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