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## Homework Statement

1) A square of area 100 cm^2 is at rest in the reference frame of O. Observer O’ moves relative to O at 0.8c along the diagonal of the square. What is the area measured by O’?

## Homework Equations

## The Attempt at a Solution

I solved it the following way:

I resolved each side of the square into 2 components- One along the diagonal and the other perpendicular to the diagonal.

Given that side of square(a)=10 cm

Component of side a along the diagonal=a(cos(45)) cm

Component of side a perpendicular to the diagonal=a(sin(45)) cm

Due to the motion of O’ only the component of a along the diagonal gets

contracted.

Contracted length= a(cos(45))[1 – v^2/c^2]^(1/2)

= 6/(2^(1/2))

Let a1 be the side of the new square w.r.t O’.

By applying Pythagoras theorem,

a1=[ {6/(2^(1/2))}^2 + { a(sin(45))}^2]^(1/2)

= sqrt(68) cm

Area of new square= (a1)^2

= 68 cm^2

But the answer the given in my book is 60 cm^2.