A small ball starts moving from A over a fixed track as shown in the figure. Surface A B has friction. From A to B the ball rolls without slipping. Surface B C is frictionless. K A , K B and K C are kinetic energies of the ball at A, B and

Question

A small ball starts moving from $A$ over a fixed track as shown in the figure. Surface $A B$ has friction. From $A$ to $B$ the ball rolls without slipping. Surface $B C$ is frictionless. $K_{A}$ , $K_{B}$ and $K_{C}$ are kinetic energies of the ball at A, B and C, respectively. Then [2006]

Smart Achievers · Accepted Answer

(1, 2, 4)

From figure given in question,

Potential energy of the ball at point A $= m g h_{A}$

Potential energy of the ball at point B = 0

Potential energy of the ball at point C $= m g h_{C}$

Total energy at point A, $E_{A} = K_{A} + m g h_{A}$

Total energy at point B, $E_{B} = K_{B}$

Total energy at point C, $E_{C} = K_{C} + m g h_{C}$

As body rolls between A and B and between B and C there is no friction. So energy should be conserved here.

By law of conservation of energy

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

$As E_{A} = E_{C}$

$K_{A} + m g h_{A} = K_{C} + m g h_{C}$

$So, if h_{A} > h_{C} \Rightarrow K_{A} < K_{C} . So option (2) is correct.$

$If h_{A} < h_{C} \Rightarrow K_{A} > K_{C}$

$Doesn't matter if h_{A} > h_{C} or h_{A} < h_{C}, we will always have K_{B} > K_{C}$

$because E_{A} = E_{B} = E_{C} . So option (1) and (4) is also correct.$

Solution

Q.
A small ball starts moving from $A$ over a fixed track as shown in the figure. Surface $A B$ has friction. From $A$ to $B$ the ball rolls without slipping. Surface $B C$ is frictionless. $K_{A}$ , $K_{B}$ and $K_{C}$ are kinetic energies of the ball at A, B and C, respectively. Then [2006]

1 $h_{A} > h_{C}; K_{B} > K_{C}$

2 $h_{A} > h_{C}; K_{C} > K_{A}$

3 $h_{A} = h_{C}; K_{B} = K_{C}$

4 $h_{A} < h_{C}; K_{B} > K_{C}$

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Solution

Q. A small ball starts moving from A over a fixed track as shown in the figure. Surface AB has friction. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and KC are kinetic energies of the ball at A, B and C, respectively. Then [2006]

1 hA>hC; KB>KC

2 hA>hC; KC>KA

3 hA=hC; KB=KC