00:01
Hi, in this question we have a rocket which is cruising past a laboratory at speed we given as 0 .650 into 10 race 2 par 6 meter per second in positive x direction just as a proton is launched with velocity.
00:25
So velocity of proton is given as 1 point of proton is given as 1 point 6 .2 into 10 raised 2 power 6 i cap plus 1 .62 into 10 raise 2 power 6 j cap and this is also in meter per second.
00:45
Now in part a of this question they're asking what is the proton speed in the lab frame? so the proton speed in lap frame can be calculated as under root 1 .1 .1 .1 .1 .5 .5 .5.
01:02
1 .62 into 10 raise 2 power 6 whole square plus 1 .62 into 10 raise 2 power 6 whole square.
01:13
As i cap and j cap both are having the same coefficients.
01:18
So we'll just take the under root and add them by taking the square of both the coefficients.
01:25
So on solving them we get the velocity of proton as 2 .29 into 10 raise 2 .6.
01:32
Meter per second.
01:34
So this is the velocity in lap frame.
01:38
Now in next part, part b, they're asking that what is the angle from the y axis of the proton speed in the lap frame? so we have to tell the angle.
01:52
So angle we can calculate as theta and this is equals to tan inverse.
02:00
So 1 .62 into 10 to power 6 divided by 1 .62 into 10 raise to power 6.
02:08
So this will be tan inverse 1...