A particle moving with kinetic energy $3 J$ makes an elastic collision (head on )with a stationary particle which has twice its mass, During impact:

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The minimum kinetic energy of the system is $1 J$

Momentum and total energy are conserved at energy instant

The ratio of kinetic energy to potential energy of the system first decreases and then increases

The maximum elastic potential energy of the systems is $2 J$

answer is A, B, C, D.

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Detailed Solution

In a head on elastic collision between two particles , the kinetic energy becomes minimum and potential energy becomes maximum at the instant when they move with a common velocity. The momentum and energy are conserved at energy instant. $\frac{1}{2} m u^{2} = 3 J m u = (m + 2 m) v = v = \frac{u}{3}$

Minimum kinetic energy of the system $\frac{1}{2} (3 m) {(\frac{u}{3})}^{2} = 1 J$

Maximum elastic potential energy of system = (3-1)J = $2 J$

During the course of collision ,due to deformation elastic potential energy of the system first increases and kinetic energy decreases , after maximum deformation the objects gradually regain their shapes and as a result elastic potential energy decreases and kinetic energy increases.

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