Two particles of masses 2m and 3m are placed at separation d on a smooth surface. They move towards each other due to mutual attractive force. Find (a) acceleration of c.m. (b) Velocity of c.m. when separation between particles becomes d//3. ( c) At what distance from the initial position of mass 2m, the particles collide.
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Given :
Mass of particle 1 (m₁) = 2m
Mass of particle 2 (m₂) = 3m
To find :
a)acceleration of center of mass
b) Velocity of center of mass
c) distance from initial position when the particle collides
Solution :
- a) As the external force applied is zero,the acceleration of center of mass will be zero
- b)Initially the particels are at rest so Vcm = 0
- As acceleration of center of mass is zero the velocity of the center of mass is constant at any instant
- c) As velocity is constant the particles will meet at their center of mass
- Xcm = (2m×0 + 3m×d) /2m+3m
= 3d/5
The particles collide at a distance of 3d/5 from the mass 2m
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(a) The acceleration of c.m is zero.
(b) The velocity of c.m. when separation between particles becomes d//3 is zero.
(c) At 3d/5 distance from the initial position of mass 2m, the particles collide.
Explanation:
(a) Acm = The external force acting on the particles
Acm = ∑F = 0
∴ Acm = 0
(b) ∵ ∑F = 0
v₁ = v₂ = constant
∴ vcm = 0
(c) The distance is given as:
Xcm = ((2m × 0) + (3m × d))/5m
Xcm = (3md)/5m
∴ Xcm = 3d/5
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