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Q.

A ball moves over a fixed track as shown in figure. From A to B the ball rolls without slipping. Surfaces BC is frictionless. $K_{A}, K_{B} a n d K_{C}$ are kinetic energies of the ball at A, B and C respectively. Then

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a

$h_{A} > h_{C} : K_{C} > K_{A}$

b

$h_{A} < h_{C} : K_{A} > K_{C}$

c

$h_{A} > h_{C} : K_{B} > K_{C}$

d

$h_{A} = h_{C} : K_{B} = K_{C}$

answer is A, B, D.

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Detailed Solution

As, $E_{A} = m g h_{A} + K_{A}$

$E_{B} = K_{B}; E_{C} = m g h_{C} + K_{C}$

Using conservation of energy

$E_{A} = E_{B} = E_{C},$

$K_{B} > K_{C}$

if $h_{A} > H_{C}; K_{C} > K_{A}$

and it $h_{A} < h_{C}; K_{C} < K_{A}$

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A ball moves over a fixed track as shown in figure. From A to B the ball rolls without slipping. Surfaces BC is frictionless. KA,KB and KC are kinetic energies of the ball at A, B and C respectively. Then