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}=m g h_{A}+K_{A}(1) $
$E_{B}=K_{B}(2) $
$E_{C}=m g h_{C}+K_{C}(3)$

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,
$m g\left(h_{A}-h_{C}\right)+\left[K_{A}-K_{C}\right]=0$
$h_{A}-h_{C}=\frac{K_{C}-K_{A}}{m g}$
From kinetic energy $K_{B}$, only translational part gets converted into potential energy at point $C$, we get $h_{C} < h_{A}$. Therefore, $K_{C} > K_{A}$