Question

A ball moves over a fixed track as shown in the figure. From $A$ to $B$, the ball rolls without slipping. Surface $BC$ is frictionless. $K_A,K_B$ and $K_C$ are kinetic energies of the ball at $A,B$ and $C$, respectively. Then

• A. $h_A>h_C;K_B>K_C$
• B. $h_A>h_C;K_C>K_A$
• C. $h_A=h_C;K_B=K_C$
• D. $h_A<h_C;K_B>K_C$

Answer 1

Answer: (A), (B), (D)

Using conservation of energy, we can write the following:
$E_A=mg{h}_A+K_A(1)$
$E_B=K_B(2)$
$E_C=mg{h}_C+K_C(3)$

Since $E_A=E_B=E_C$, we have the following:
$K_B>K_A$
$K_B>K_C$
From Eqs. (1) and (3), we get
$E_A-E_C=0$
That is,
$mg\left(h_A-h_C\right)+\left[K_A-K_C\right]=0$
$h_A-h_C=\frac{K_C-K_A}{mg}$
From kinetic energy $K_B$, only translational part gets converted into potential energy at point $C$, we get $h_C<h_A$. Therefore, $K_C>K_A$