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Hello IITIAN!! I HAVE A PROBLEM AND I HOPE YOU WILL RESOLVE IT..A BALL OF 1 KG STRIKES A WALL WITH VELOCITY 1 m/s AT AN ANGLE OF 60° WITH THE WALL AND REFLECTS AT THE SAME ANGLE. IF IT REMAINS IN CONTACT WITH WALL FOR 0.1 SEC,THEN THE FORCE IS ?MY QUESTION IS NOT COMPLETE YET.... I REQUEST YOU TO ANSWER IT BY DIVIDING THE ANGLES INTO ITS COMPONENTS AND BY USING DIAGRAM.. (ANS.10√3 N)PLZ EXPLAIN WITH FULL METHOD. THANK YOU

Ashok Mishra , 6 Years ago
Grade 9
anser 1 Answers
Khimraj

Last Activity: 6 Years ago

I can’t draw image here but try to explain well
force is given by M*\Delta v/\Delta t
initial velocity = sin60\hat{i} – cos60\hat{j}
final velocity = -sin60\hat{i} – cos60\hat{j}
\Delta v = 2sin60\hat{i}
|\Delta v| = \sqrt{3}
\Delta t = 0.1
then F = 1*\sqrt{3}/0.1 = 10\sqrt{3} N
Hope it clears.

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