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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A spherical ball& A of mass 4 kg, moving along a straight line...

A spherical ball& AA of mass 44 kg, moving along a straight line strikes another spherical ball BB of mass 11 kg at rest. After the collision, AA and BB move with velocities v1ms–1{v}_{1}ms^{– 1} and v2ms–1{v}_{2}ms^{– 1} respectively making angles of 30°30 ° and 60°60 ° with respect to the original direction of motion of AA. The ratio v1v2\dfrac{v1}{v2}& will beA spherical ball& A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest.

Options

  1. A34\dfrac{\sqrt{3}}{4}
  2. B43\dfrac{4}{\sqrt{3}}
  3. C13\dfrac{1}{\sqrt{3}}
  4. D3\sqrt{3}

Correct answer

A. 34\dfrac{\sqrt{3}}{4}

Step-by-step solution

A spherical ball& A of mass 4 kg, moving along a straight line strikes another spherical ball B of mass 1 kg at rest.

Applying the law of conservation of linear momentum along a direction perpendicular to the direction of motion (i.e. along the y-axis), we get.

0+0=4v1sin⁡30°–v2sin⁡60°0 + 0 = 4{v}_{1}\sin 30 ° – {v}_{2}\sin 60 °

4v1sin⁡30°=v2sin⁡60°4{v}_{1}\sin 30 ° = {v}_{2}\sin 60 °

\dfrac{v1}{v2} = \dfrac{\text\sin6{0}^{o}}{4\text\sin3{0}^{o}} = \dfrac{3}{4}
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