I have reworked using a ruler length of L instead of 1m. I get:
1. The right end of the ruler passes the observer (in the observer's frame) (0,0)
2. The left end of the ruler passes the observer (in the observer's frame) (0,L/(u*gamma))
3. The observer passes the right end of the ruler (in...
Homework Statement
A meter ruler moves at velocity u to the right past a stationary observer. The observer is at (0,0) in his rest frame. Give the (x,t) co-ordinates of the following events.
1. The right end of the ruler passes the observer (in the observer's frame)
2. The left end of the...
Homework Statement
I know that if you have 2 observers, "A", one at rest and the other "B" moving wrt "A" and if the moving observer shoots a projectile, then we can calculate the velocity of the projectile wrt "A" using the standard equation shown below.
I was thinking what equation would...
Yes!
Exactly!
Here are my current thoughts.
Taking down as positive:
For 8Kg weight, where a is the acceleration of the 8Kg:
8a=-T1+8g
For 14Kg weight where b is the acceleration of the 14Kg:
14b=14g-T2
For Pulley (since the tension on both sides of the string acts down)...
Oops. Seems I skipped a step! Thanks.
Intuitively
8a=T1-8g
14a=14g-T2
T1=T2
=> (8+14)a= 14g-8g
which gives the correct answer after a bit of algebra.
ONE FURTHER QUESTION
If the down direction is taken as positive, shouldn't these equations be...
(T1 act upwards
8a=-T1+8g
14a=14g-T2
T1=-T2...
Homework Statement
A 14kg mass is attached to one side of a vertical pulley and an 8kg to the other.
The 14kg mass is 5m above the ground. The 8kg is just resting on the ground. The pulley is frictionless and weightless.
Find the velocity of the 14kg mass just before it hits the...