Three identical particles A, B and C (each of mass m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is 135⁰ as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,
The speed of the particle ‘C’ is $\frac{n J}{7 m}$ . The value of n is

Question

Three identical particles A, B and C (each of mass m) lie on a smooth horizontal table. Light inextensible strings which are just taut connect AB and BC and ABC is 135⁰ as shown in diagram. An impulse J is applied to the particle C in the direction BC for a very short time interval. Then just after applying impulse J,

The speed of the particle ‘C’ is $\frac{n J}{7 m}$ . The value of n is

Infinity Learn · Accepted Answer

The external impulse applied to C causes both strings to jerk exerting internal impulses J₁ and J₂

$\begin{array}{l} V_{2} = v_{1 x} - - - - - - - (1) \\ J_{1} \cos 45^{\circ} - J_{2} = {mv}_{1 x} - - - - - - - - - (3) \\ J_{1} \cos 45^{\circ} = {mv}_{1 y} - - - - - - - (4) \\ J - J_{1} = mv - \dots . . . . . - (5) \end{array}$

Also velocities of B&C along BC are are equal i.e.,

$v_{1 y} \cos 45^{0} + v_{1 x} \cos 45^{0} = v - - - - - (6)$
After solving we get, $v_{2} = v_{1 x} = \frac{\sqrt{2} J}{7 m}$

$v = \frac{3 J}{7 m}; v_{1 y} = \frac{2 \sqrt{2} J}{7 m}; V_{A} = \frac{\sqrt{2} J}{7 m}, V_{B} = \frac{\sqrt{10} J}{7 m}, J_{1} = \frac{3 J}{m} a n d J_{2} = \frac{\sqrt{2} J}{7}$

Law of conservation of momentum and it's applications

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