AP EAPCET 23-Aug-2021 Shift 2 Paper

Given, length of pendulum, l=0.5mSpeed at lowest point, v1=6m∕s
Angle made by string with vertical, θ=60°
In Δ‌OBC,

OC

OB

=cos‌θ=cos‌60°

l−h

l

=cos‌60°
⇒h=l(1−cos‌60°)
⇒h=0.5(1−

1

2

)=0.25m
Using law of conservation of mechanical energy, we get
Ki+Ui=Kf+Uf
⇒

1

2

mv12+0=

1

2

mv22+mgh
[Taking initial potential energy at ground is equal to zero.]
⇒

1

2

mv22=

1

2

mv12−mgh
⇒v22=v12−2gh
Substituting the values, we get
v2=√(6)2−2×10×0.25 =√31m∕s
Hence, the final velocity at an angle 60° is √31m∕s.