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  • Solve it, - Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B ha - Work Energy and Power - JEE Main

Three blocks A, B and C are lying on a smooth horizontal surface, as shown in the figure. A and B have equal masses, m while C has mass M. Block A is given an inital speed v towards B due to which it collides with B perfectly inelastically. The combined mass collides with C, also perfectly inelastically \frac{5}{6}^{th} of the initial kinetic energy is lost in whole process. What is value of M/m?

 

  • Option 1)

    5

  • Option 2)

    2

  • Option 3)

    4

  • Option 4)

    3

Answers (1)

best_answer

 

Inelastic Collision -

Law of conservation of momentum hold good but kinetic energy is not conserved

- wherein

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}\neq \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}

m_{1},m_{2}: masses

u_{1},v_{1}: initional \: and \: final\: velocities\: of \: mass \: m_{1}

u_{2},v_{2}: initional \: and \: final\: velocities\: of \: mass \: m_{2}

 

 

 


Option 1)

5

Option 2)

2

Option 3)

4

Option 4)

3

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