Two particles of mass $m$ each are tied at the ends of a light string of length $2a$ . The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $'a'$ from the centre $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$ . As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$ , is
Two particles of mass $m$ each are tied at the ends of a light string of length $2a$ . The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $'a'$ from the centre $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$ . As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$ , is
- A
$\frac{F}{{2m}}\,\frac{a}{{\sqrt {{a^2} - {x^2}} }}$
- B
$\frac{F}{{2m}}\,\frac{x}{{\sqrt {{a^2} - {x^2}} }}$
- C
$\frac{F}{{2m}}\,\frac{x}{a}$
- D
$\frac{F}{{2m}}\,\frac{{\sqrt {{a^2} - {x^2}} }}{x}$
Similar Questions
Three forces starts acting simultaneously on a particle moving with velocity $\vec v.$ These forces are represented in magnitude and direction by the three sides of a triangle $ABC$ (as shown). The particle will now move with velocity
Three forces starts acting simultaneously on a particle moving with velocity $\vec v.$ These forces are represented in magnitude and direction by the three sides of a triangle $ABC$ (as shown). The particle will now move with velocity
- [AIEEE 2003]
Define impulse of force. Time derivative of momentum gives which physical quantity ?
Define impulse of force. Time derivative of momentum gives which physical quantity ?
See Figure given below. A mass of $6 \;kg$ is suspended by a rope of length $2 \;m$ from the ceiling. A force of $50\; N$ in the horizontal direction is applied at the midpoint $P$ of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take $g = 10 \;m s^{-2}$). Neglect the mass of the rope.
See Figure given below. A mass of $6 \;kg$ is suspended by a rope of length $2 \;m$ from the ceiling. A force of $50\; N$ in the horizontal direction is applied at the midpoint $P$ of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take $g = 10 \;m s^{-2}$). Neglect the mass of the rope.
What is Free body diagram ?
What is Free body diagram ?
A mass of $10 \,kg$ is suspended vertically by a rope of length $5 \,m$ from the roof. A force of $30 \,N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}\left(x \times 10^{-1}\right)$. The value of $x$ is ................
$\text { (Given } g =10 \,m / s ^{2} \text { ) }$
A mass of $10 \,kg$ is suspended vertically by a rope of length $5 \,m$ from the roof. A force of $30 \,N$ is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta=\tan ^{-1}\left(x \times 10^{-1}\right)$. The value of $x$ is ................
$\text { (Given } g =10 \,m / s ^{2} \text { ) }$
- [JEE MAIN 2022]