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  •  After perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle would be Option: 1</stro

 After perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle would be 

Option: 1 80

Option: 2 70

Option: 3 90

Option: 4 45

Answers (1)

 

 

Perfectly elastic oblique collision -

  • Let two bodies moving as shown in figure.

                                                         

By law of conservation of momentum

Along x-axis-

m_1u_1+m_2u_2 = m_1v_1cos\theta+m_2v_2cos\phi ….. (1)

Along y-axis-

0 = m_1v_1sin\theta-m_2v_2sin\phi                            …..(2)

By law of conservation of kinetic energy

                       \frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}       ….(3)

And In Perfectly Elastic Oblique Collision

Value of e=1 

So along line of impact (here along in the direction of v_2)

We apply e=1 

And  we get e=1=\frac{v_2-v_1cos(\theta+\phi)}{u_1cos\phi-u_2cos\phi}     ….. (4)

So we solve these  equations (1),(2),(3),(4) to get unknown.

  • Special condition

if\ m_1 = m_2\ and\ u_2 = 0

Then, from equation (1), (2) and (3)

We get,  \theta+\phi=\frac{\pi }{2}

i.e; after perfectly elastic oblique collision of two bodies of equal masses (if the second body is at rest), the scattering angle  \theta+\phi  would be  90^{0}.

So, answer is - 

90^{0}

Posted by

Sumit Saini

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