This set of Engineering Physics Multiple Choice Questions & Answers (MCQs) focuses on “Law of Addition of Velocities”.

1. Two particles approach each other with a speed 0.8 c with respect to the laboratory. Their relative speed is ___________

a) 0.912 c

b) 0.95 c

c) 0.975 c

d) 0.85 c

View Answer

Explanation: ux = \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\)

= 0.6c/1.64

= 0.975 c.

2. What would be the mass of the body at rest, if it explodes into two objects of mass 2 kg each moving with a speed 0.8c relative to the original body?

a) 4 Kg

b) 5.4 Kg

c) 6.2 Kg

d) 6.7 Kg

View Answer

Explanation: E

_{0}= mc

^{2}= γm

_{1}c

^{2}+ γm

_{2}c

^{2}

E

_{0}= \(\frac{2mC^2}{\sqrt{\frac{1-v^2}{C^2}}}\)

m = E/c

^{2}

m = \(\frac{2 X 2}{\sqrt{1-0.64}}\)

m = 6.67 Kg.

3. In the case v << c, Lorentz transformation is the same as ____________

a) Einstein’s transformation

b) Galilean transformation

c) Maxwell’s transformation

d) Planck’s transformation

View Answer

Explanation: At v << c, u

_{x}= u

^{’}

_{x}+ v

U

_{y}= u

_{y}

^{’}, which is the same as Galilean transformations.

4. When a particle is moving with a velocity of light c relative to S, its velocity as observed by an observer in the frame S^{’} is _____________

a) Zero

b) 0.5 c

c) 0.75 c

d) c

View Answer

Explanation: As the particle moves with a speed c, u = c.

Therefore, u

^{’}= \(\frac{C-v}{1-\frac{v}{C}} \) = c

Hence, the velocity as observed by the observer in frame S’ is still c.

5. Lorentz transformations are based on the principle of consistency of the velocity of light.

a) True

b) False

View Answer

Explanation: In Lorentz transformation, if the particle is moving with the velocity of light in a frame, it’s velocity as observed from another frame remains c. This proves that Lorentz transformation is based on the principle of consistency of light.

6. In Lorentzian relativity, if two events are simultaneous for one observer, they will be simultaneous for all other observers as well.

a) True

b) False

View Answer

Explanation: In Newtonian and Galilean relativity, the above-stated condition is followed. However, in Lorentzian relativity, if two events occur simultaneously at the instant t at x1 and x2, then the two events are not simultaneous.

7. For u^{’} < c and v < c the equation becomes ____________

a) u = \(\frac{u^{‘}+v}{\frac{u^{‘}v}{c^2}}\)

b) u = \(\frac{u^{‘}+v}{1+u^{‘}v}\)

c) u = 1+\(\frac{u^{‘}v}{c^2} \)

d) u = u’ + v

View Answer

Explanation: In this case, u’< c and v < c

Therefore, u’v < c

^{2}

u’v/c

^{2}< 1

Thus the equation u = \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\) changes to

u = u’ + v.

8. From the graph, what should be the velocity as observed from frame S?

a) 0.6 c

b) 0.7 c

c) 0.8 c

d) 0.9 c

View Answer

Explanation: Now, u

_{x}= \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\)

Here, u’ = 0.4 c and v = 0.6 c

Therefore, u

_{x}= c/1 + 0.24

= 0.8 c.

9. Two particles approach each other with a velocity of 0.9 c. What is their relative velocity as observed by A?

a) 0.9 c

b) 0

c) 0.99 c

d) 0.94 c

View Answer

Explanation: u’ = 0.9c, v = 0.9c

u

_{x}= \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\)

= 1.8c/1.81

= 0.99 c.

10. An atom A, moving relative to the observer, with velocity 2 X 10^{8}m/s emits a particle B which moves with a velocity of 2.8 X 10^{8}m/s with respect to the atom. The velocity of the emitter particle relative to the scientist is _____________

a) 0.8 X 10^{8}m/s

b) 2.4 X 10^{8}m/s

c) 3 x 10^{8}m/s

d) 2.95 X 10^{8}m/s

View Answer

Explanation: Now we know, u

_{x}= \(\frac{u^{‘}+v}{1+\frac{u^{‘}v}{c^2}}\)

Here, u’ = 2.8 X 10

^{8}m/s

u = 2 X 10

^{8}m/x

u

_{x}= ?

u

_{x}= 4.8/ (1 + 5.6/3) X 10

^{8}m/s

= 2.95 X 10

^{8}m/s.

**Sanfoundry Global Education & Learning Series – Engineering Physics.**

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