elastic bouncing

[kavlas]

Suppose a cannon fires horizontaly a perfectly elastic ball against a perfectly elastic wall, distant x, with speed v and at the same time starts rushing against the wall with speed u.

 

Will the time taken for the ball to hit the wall and come back to the cannon be given by the following formula:

 

[math] t=\frac{2x}{v+u}[/math]??

[Mr Skeptic]

Yes. I probably solved this differently than you did, but it turns out to be equivalent to the cannon and cannonball moving towards each other.

[kavlas]

Yes. I probably solved this differently than you did, but it turns out to be equivalent to the cannon and cannonball moving towards each other.

 

Thank you. Now i have the following question:

 

Suppose the distance between the cannon and the wall is c kilometres ( c is the speed of light) and the ball is fired against the wall with c speed ,while the cannon is rushing against the wall with c/2 speed.After what time the ball will meet the cannon??Is the above formula applicable in this case too??

[Mr Skeptic]

Thank you. Now i have the following question:

 

Suppose the distance between the cannon and the wall is c kilometres ( c is the speed of light)

 

Would that be 1 light-second then? Or a light-hour?

 

and the ball is fired against the wall with c speed ,while the cannon is rushing against the wall with c/2 speed.After what time the ball will meet the cannon??Is the above formula applicable in this case too??

 

The answer will of course depend on the frame of reference (which will change the apparent speed of the cannon but not the "cannonball", and also the distance). But yes, after adjusting the distance and velocity changes (if you use a different reference frame), you should be able to use that formula.

[kavlas]

Would that be 1 light-second then? Or a light-hour?

 

 

 

The answer will of course depend on the frame of reference (which will change the apparent speed of the cannon but not the "cannonball", and also the distance). But yes, after adjusting the distance and velocity changes (if you use a different reference frame), you should be able to use that formula.

 

 

So to the standing observer a distance x from where he stands an event has taken place ( a cannonball has hit a wall 300000kilometers (c klm) away and come back to the cannon) after t time .

 

 

 

 

That time t =[math]\frac{2c}{C/2 +C}[/math] = 4/3 sec.

 

Now to the observer on the cannon the same event has taken place with new coordinates of time t' and x'=0.

 

And from the Lorentz transformations: [math]x' =\frac{x-vt}{\sqrt{1-\frac{v^2}{c^2}}}[/math] we have:

 

x = vt since x'=0.But v=c/2 and t=4/3 sec and so x=[math]\frac{4c}{6}[/math]klm.

 

And from the trasformation:

 

 

[math]t'=\frac{t-\frac{vx}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}[/math] we can get t' by putting v=c/2klm/sec and x=4c/6 klm.

 

That time will be shorter than t.

 

Now is trere a checking procedure for the above??

[Mr Skeptic]

From an observer on the cannon, v=0 for the cannon, and the cannonball still moves at c.

[kavlas]

From an observer on the cannon, v=0 for the cannon, and the cannonball still moves at c.

 

 

Then we will have : t' =t

 

Besides v is the speed of his reference frame

Edited August 16, 2010 by kavlas