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BITSAT 2010
Exam
BITSAT
Year
2010
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

BITSAT 2010 · Physics · Center of Mass, Momentum and Collision

A spring is compressed between two toy carts of mass m_ 1 and m_ 2...

A spring is compressed between two toy carts of mass m1\mathrm{m}_{1} and m2\mathrm{m}_{2}. When the toy carts are released, the springs exert equal and opposite average forces for the same time on each toy cart. If v1v_{1} and v2\mathrm{v}_{2} are the velocities of the toy carts and there is no friction between the toy carts and the ground, then :

Options

  1. Av1/v2=m1/m2\mathrm{v}_{1} / \mathrm{v}_{2}=\mathrm{m}_{1} / \mathrm{m}_{2}
  2. Bv1/v2=m2/m1\mathrm{v}_{1} / \mathrm{v}_{2}=\mathrm{m}_{2} / \mathrm{m}_{1}
  3. Cv1/v2=−m2/m1\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{2} / \mathrm{m}_{1}
  4. Dv1/v2=−m1/m2\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{1} / \mathrm{m}_{2}

Correct answer

C. v1/v2=−m2/m1\mathrm{v}_{1} / \mathrm{v}_{2}=-\mathrm{m}_{2} / \mathrm{m}_{1}

Step-by-step solution

Applying law of conservation of linear momentum m1v1+m2v2=0m1m2=−v2v1\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0\frac{\mathrm{m}_{1}}{\mathrm{ m}_{2}}=-\frac{\mathrm{v}_{2}}{\mathrm{v}_{1}} or v1v2=−m2m1\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=-\frac{\mathrm{m}_{2}}{\mathrm{ m}_{1}}
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