Two blocks on a frictionless horizontal surface are on a collision course. one block with mass 0.6 kg moves at 0.8 m/s to the right and collides with a 1.2 kg mass at rest and the two masses bounce off each other elastically. find the final velocities of the two masses after the collision.

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aristocles aristocles · Answer 1 · 2018-08-03T03:40:57+02:00

First we can say that since there is no external force on this system so momentum is always conserved.

[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]

[tex]0.6*0.8 + 1.2*0 = 0.6*v_{1f} + 1.2*v_{2f}[/tex]

[tex]0.48= 0.6*v_{1f} + 1.2*v_{2f}[/tex]

[tex]0.8 = v_{1f} + 2v_{2f}[/tex]

now by the condition of elastic collision

[tex]v_{2f} - v_{1f} = 0.8 - 0[\tex]

now add two equations

[tex]3*v_{2f} = 1.6[/tex]

[tex]v_{2f} = 0.533 m/s[/tex]

also from above equation we have

[tex]v_{1f} = -0.267 m/s[/tex]

So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.