Question

A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. If surface BC is friction less and $K_A$, $K_B$ 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

The correct option is A $h_A>h_C;K_B>K_C$

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

To find the relation between them

Solution:

From the figure,

Energies at the points A, , C will be

$E_A=mg{h}_A+K_A.......\left(i\right)E_B=K_B...........\left(ii\right)E_C=mg{h}_C+K_C............\left(iii\right)$

Using conservation of energy

$E_A=E_B=E_C$

equating eqn(i)and (iii), we get

$mg{h}_A+K_A=mg{h}_C+K_C⟹mg(h_A-h_C)=K_C-K_A⟹h_A-h_C=\frac{K_C-K_A}{mg}⟹h_A=\frac{K_C-K_A}{mg}+h_C⟹h_A>h_C$

And by comparing eqn(i) , (ii) and (iii), we find that

$K_B>K_C,K_B>K_A$