Learn to Create a Robotic Device Using Arduino in the Free Webinar. Register Now
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Free webinar on Robotics (Block Chain) Learn to create a Robotic Device Using Arduino
30th Jan @ 5:00PM for Grade 1 to 10
Plz experts answer my question as fast as possible plzzz..... Plz experts answer my question as fast as possible plzzz.....
Answer is 5ucos(theeta)/2.Initial and final momentum along y direction is zeroApplying conservation of momentum along x direction just before and just after explosion :4mucos(theeta) = -mucos(theeta) + 2mv=> v = 5ucos(theeta)/2Note that the whole mass of particle before explosion is 4m and at the highest point of it`s trajectory, it has only x-component of velocity I.e. vcos(theeta)So initial momentum along x direction is (4m)ucos(theeta). ( Note that we have assumed right direction as +ve and left direction as negative)For final momentum of system, the particle of mass m which retraces it`s path will have velocity of -ucos(theeta) ( negative sign because it retraces it`s path, i.e. it`s final velocity is along left direction).Further let the velocity of 2m be v.So final momentum along x-axis is -mucos(theeta) + 2mvNow equating initial and final momentum along x-axis4mucos(theeta) = -mucos(theeta) + 2mv=> v = 5mucos(theeta)/2.So final velocity of 2m is 5mucos(theeta)/2 along right direction.
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -