Two particles , of masses M and 2M , moving as shown , with speeds of 10 m/s and 5 m/s ,

collide elastically at the origin.After the collision , they move along the indicated directions with

speeds v_{1} and v_{2}, respectively. The values of v_{1} and v_{2} are nearly :

  • Option 1)

    6.5 m/s and 6.3 m/s

  • Option 2)

    3.2 m/s and 6.3 m/s

  • Option 3)

    6.5 m/s and 3.2 m/s

  • Option 4)

    3.2 m/s and 12.6 m/s

 

Answers (1)

 

Elastic Collision in 2 dimension -

\fn_jvn \vec{P_{i}}= \vec{P_{f}}

m_{1}v_{0}\: \: \hat{i}= \left ( m_{1}v_{1} \cos \Theta + m_{2}v_{2} \cos \beta \right )\hat{i}+\left ( m_{1}v_{1} \sin \Theta - m_{2}v_{2} \sin \beta \right )\hat{j}

- wherein

 

 

u_1                  u_{1x}=10\times \frac{\sqrt3}{2}\hat{i}

                      u_{1y}=10\times \frac{1}{2}\hat{j}=-5\hat{j}                                 M_1=M

u_2                u_{2x}= \frac{5}{\sqrt2}\hat{i}

                    u_{2y}= \frac{5}{\sqrt2}\hat{j}                                                       M_2=2M

v_1              v_{1x}=v_1\times \frac{\sqrt3}{2}\hat{i}

                   v_{1y}=v_1\times \frac{1}{2}\hat{j}

v_2               v_{2x}=v_2\times \frac{1}{\sqrt2}\hat{i}

                   v_{2y}= \frac{v_2}{\sqrt2}(\hat{-j})

\Delta P_x=0=>m\times (\frac{10\sqrt3}{2})+(2m\times \frac{5}{\sqrt2})= 2m(\frac{\sqrt3}{2})v_1+m(\frac{v_2}{\sqrt2})

                  => 5\sqrt3+5\sqrt2=\sqrt3v_1+\frac{v_2}{\sqrt2}................................(1)

\Delta P_y=0=>-m\times (5)+(2m\times \frac{5}{\sqrt2})= 2m(\frac{v_1}{2})-m(\frac{v_2}{\sqrt2})

                   =>5\sqrt2-5=v_1-\frac{v_2}{\sqrt2}............................................(2)

On adding (1) and (2)

5(\sqrt3-1)+10\sqrt2=(\sqrt3+1)v_1

=> v_1=\frac{5(\sqrt3-1+10\sqrt2)}{\sqrt3+1}\approx 6.5

=> v_2\approx 6.3

 

 

 


Option 1)

6.5 m/s and 6.3 m/s

Option 2)

3.2 m/s and 6.3 m/s

Option 3)

6.5 m/s and 3.2 m/s

Option 4)

3.2 m/s and 12.6 m/s

Exams
Articles
Questions