SATHEE: Work Energy and Power - Result Question 53

Work Energy and Power - Result Question 53

56. Two spheres $A$ and $B$ of masses $m_{1}$ and $m_{2}$ respectively collide. $A$ is at rest initially and $B$ is moving with velocity $v$ along $x$ -axis. After collision $B$ has a velocity $\frac{v}{2}$ in a direction perpendicular to the original direction. The mass $A$ moves after collision in the direction.

(a) Same as that of $B$

[2012]

(b) Opposite to that of $B$

(d) $θ = \tan^{- 1} (- 1 / 2)$ to the $x$ -axis

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Answer:

Correct Answer: 56. (d)

Solution:

[Before collision]

[After collision] According to law of conservation of linear momentum along $x$ -axis, we get

$\Rightarrow m_{1} (0) + m_{2} v = m_{1} v^{'} \cos θ$

$\Rightarrow \cos θ = \frac{m_{2} v}{m_{1} v^{'}}$

Similarly, law of conservation of linear momentum along $y$ -axis, we get

$\Rightarrow m_{1} (0) + m_{2} (0) = m_{1} v^{'} \sin θ + m_{2} (\frac{v}{2})$

$\Rightarrow \sin θ = \frac{m_{2} v}{2 m_{1} v^{'}}$

From equations, (i) and (ii),

$\tan θ = - (\frac{m_{2} v}{2 m v^{'}}) \times (\frac{m_{1} v^{'}}{m_{2} v}) = - \frac{1}{2}$

$θ = \tan^{- 1} (\frac{- 1}{2})$ to the $x$ -axis.

Let A moves in the direction, which makes an angle $θ$ with initial direction i.e.,

$\tan θ = \frac{v_{y}}{v_{x}} = \frac{m_{2} v}{2 m_{1}} / \frac{m_{2} v}{2 m_{1}}$

$\tan θ = \frac{1}{2}$

$\Rightarrow θ = \tan^{- 1} (\frac{1}{2})$ to the $x$ -axis.

Work Energy and Power - Result Question 53

Answer:

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Work Energy and Power - Result Question 53

56. Two spheres A and B of masses m1 and m2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v2 in a direction perpendicular to the original direction. The mass A moves after collision in the direction.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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