Question

Two particles of masses m and 2m are connected by a string of length L and placed at rest over a smooth
horizontal surface. The particles are then given velocities as indicated in the figure shown. The tension developed in the string will be?
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Answer 1

Q]____?

Ans : T = ( 8 mv^2 ) /3L N

Explanation:

To solve this prob first we have to find centre of mass (C.O.M )

rᴄᴍ = ( m1 r1 + m2 r2 ) / m1 + m2

Now considering origin on 2m the Coordinates of masses will be :

2m = ( 0,0 )

and m = ( 2,0 )

[ That is, m1 = 2m, m2 = m

r1= 0, r2 = L ]

∴ xᴄᴍ = ( m1 v1 + m2 r2) /m1 +m2

∴ xᴄᴍ = ( m × L ) / 2m

∴ xᴄᴍ = L / 3

Now finding Velocity of C.O.M,

vc = ( m1 v1 + m2 v2 ) / m1 + m2

vc = [ ( 2m × v ) + m(-v)] / 3m

...[considering downward velocity as +ve and upward as -ve ]

vc = mv / 3m

vc = v/3

Relative velocity wrt C.O.M

Vrel = v - v/3

Vrel = 2v / 3

∴ Velocity of mass A = 2 v/3

Now finding tension;

T = mv^2 / R

T = 2m [ (2v/3 )^2 ] / ( L/3 )

T = ( 2m × 4^2 × 3 ) / 9 × L

T = ( 8 mv^2 ) /3L (N)

Thus, the tension developed in the string will be

T = ( 8 mv^2 ) /3L N !

Answer 2

HOPE THIS HELPS YOU....