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Grade: 9
        an object of 2kg is suspended using light rope of length 50 cm . It is taken through 60 degrees in 1 direction and released with the intention of giving it a pendulum motion . before the ball could reach the maximum height on the other side , the rope breaks when the ball has gone through 26 degrees at other side . As the ball undergoes free fall , calculate the speed of the ball 2 sec after the rope breaks . Express your answer in M/SGiven-Sin26=0.45Cos26=0.9
one year ago

Answers : (1)

Ilamparithi
19 Points
							
So here first we will be conserving mechanical energy
We will take the bottom point for reference (for gravitational potential energy)
Initial mech energy =final mech energy.      Mgl(1-cos60)=mgl(1-cos26)+1/2mv^2.
Substituting values we get v as 0.7m/s.   Now the horizontal component of v is vcos26 and vertical is vsin26.
Horizontal component never changes . Magnitude of vertical component after 2sec is .7-10*2 (v=u+at). By this vertical velocity is 19.3 horizontal is 0.63 . We get nett velocity as root of 19.3^2+0.63^2  = 19.3102 . I don't know if and is correct but all equations are right .
one year ago
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Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


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