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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NEET 2011 · Physics · Center of Mass, Momentum and Collision

A mass m moving horizontally (along the x -axis) with velocity v...

A mass mm moving horizontally (along the xx-axis) with velocity vv collides and sticks to mass of 3m3 m moving vertically upward (along the yy-axis) with velocity 2v2 v. The final velocity of the combination isA mass m moving horizontally (along the x -axis) with velocity v collides and sticks to mass of 3 m moving vertically

Options

  1. A14vi^+32yj^\frac{1}{4} v \hat{\mathbf{i}}+\frac{3}{2} y \hat{\mathbf{j}}
  2. B13vi^+23vj^\frac{1}{3} v \hat{\mathbf{i}}+\frac{2}{3} v \hat{\mathbf{j}}
  3. C23vi^+13j^\frac{2}{3} v \hat{\mathbf{i}}+\frac{1}{3} \hat{\mathbf{j}}
  4. D32vi^+14vj^\frac{3}{2} v \hat{\mathbf{i}}+\frac{1}{4} v \hat{\mathbf{j}}

Correct answer

A. 14vi^+32yj^\frac{1}{4} v \hat{\mathbf{i}}+\frac{3}{2} y \hat{\mathbf{j}}

Step-by-step solution

From the law of conservation of linear momentum A mass m moving horizontally (along the x -axis) with velocity v collides and sticks to mass of 3 m moving vertically mv+(3m)(2v)=(4m)v′mvi^+6mvj^=4mv′v′=14vi^+64vj^y′=14vi^+32vj^\begin{aligned} m v+(3 m)(2 v) & =(4 m) v^{\prime} \\ m v \hat{i}+6 m v \hat{j} & =4 m v^{\prime} \\ v^{\prime} & =\frac{1}{4} v \hat{i}+\frac{6}{4} v \hat{j} \\ y^{\prime} & =\frac{1}{4} v \hat{i}+\frac{3}{2} v \hat{j} \end{aligned}
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