Question
This seems like a pretty straightforward question but it's just not making sense at the moment.
A meter stick is projected into space at so great a speed that its length appears to have contracted only 50 cm. How fast is it going?
So you want the velocity, v. (in terms of c).
So I used this formula, the Lorentz contraction formula and got:
L=L'(1+v^2/c^2)^(1/2) L being a meter stick so 100 cm and L' being the apparent L-50 = 50 cm.
100 = 50(1+v^2/c^2)^1/2
2=(1+v^2/c^2)1/2
4=(1+v^2/c^2)
3=v^2/c^2
v= square root of 3 times c ??
However the true answer is .866c. Which I remember to be root 3 over 2.
A meter stick is projected into space at so great a speed that its length appears to have contracted only 50 cm. How fast is it going?
So you want the velocity, v. (in terms of c).
So I used this formula, the Lorentz contraction formula and got:
L=L'(1+v^2/c^2)^(1/2) L being a meter stick so 100 cm and L' being the apparent L-50 = 50 cm.
100 = 50(1+v^2/c^2)^1/2
2=(1+v^2/c^2)1/2
4=(1+v^2/c^2)
3=v^2/c^2
v= square root of 3 times c ??
However the true answer is .866c. Which I remember to be root 3 over 2.