Question
Two identical beads of $$m = 100\,gram$$ are connected by an inextensible massless string can slide along the two arms $$AC$$ and $$BC$$ of a rigid smooth wire frame in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of $$0.1\,m$$ is $$8x \times {10^{ - 3}}J.$$ Find the value of $$x.\left( {g = 10\,m/{s^2}} \right)$$
Two identical beads of $$m = 100\,gram$$ are connected by an inextensible massless string can slide along the two arms $$AC$$ and $$BC$$ of a rigid smooth wire frame in a vertical plane. If the system is released from rest, the kinetic energy of the first particle when they have moved by a distance of $$0.1\,m$$ is $$8x \times {10^{ - 3}}J.$$ Find the value of $$x.\left( {g = 10\,m/{s^2}} \right)$$
A.
8
B.
6
C.
9
D.
11
Answer :
8
Solution :
Let $$m$$ be the mass of the beads.
By energy conservation
$$\eqalign{ & mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2\,.....\left( {\text{i}} \right) \cr & \frac{{{v_1}}}{{{v_2}}} = \frac{4}{3}\,......\left( {{\text{ii}}} \right) \cr & {v_1} = \frac{{4\sqrt 2 }}{5},{v_2} = \frac{{3\sqrt 2 }}{5} \cr & K.E. = \frac{1}{2} \times 100 \times {10^{ - 3}} \times \frac{{16 \times 2}}{{25}} = 64 \times {10^{ - 3}}\,J \cr & x = 8. \cr} $$
Let $$m$$ be the mass of the beads.
By energy conservation
$$\eqalign{ & mgh = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2\,.....\left( {\text{i}} \right) \cr & \frac{{{v_1}}}{{{v_2}}} = \frac{4}{3}\,......\left( {{\text{ii}}} \right) \cr & {v_1} = \frac{{4\sqrt 2 }}{5},{v_2} = \frac{{3\sqrt 2 }}{5} \cr & K.E. = \frac{1}{2} \times 100 \times {10^{ - 3}} \times \frac{{16 \times 2}}{{25}} = 64 \times {10^{ - 3}}\,J \cr & x = 8. \cr} $$