Answer to Question #178517 in Mechanics | Relativity for Zeke Miller

Question

A ball of mass m is suspended from a rope of length l and at rest is a height h above the ground. The ball is pulled back to an angle of θ 1 and released. When the rope makes an angle of θ 2 and the ball is on its way up, the string breaks. Given where the ball lands with respect to its rest position, determine the height h the ball is above the ground when in the rest position.

m =5.0 kg

θ 1 = 80°

θ 2 = 50°

l = 1.5 m

Δdx = 3.5 m

Answer 1

Potential energy of the body when the ball is held at $(\theta_1=80\degree)=mgh_1$

Potential energy of the body just before the string breaks ($\theta_2=50\degree)=mgh_2$

$h_1=l(1-cos\theta_1)=1.23\space m\\ h_2=l(1-cos\theta_2)=0.535\space m$R

Applying conversation of energy,

$\Rightarrow mgh_1=mgh_2+\dfrac{1}{2}mv^2\\\Rightarrow v^2=2g(h_1-h_2)$

$\Rightarrow v =\sqrt{2g(h_1-h_2)}$

$\Rightarrow v=3.69\space m/s$

$\Delta dx=3.5\space m\\\Rightarrow\Delta dx=\dfrac{v}{g}\sqrt{v^2+2gH}$

$\Rightarrow H=\dfrac{1}{2g}\{(\Delta dx)^2\dfrac{g^2}{v^2}-v^2\}$

$\Rightarrow H=3.71\space m$