On a frictionless, horizontal air track. the glider has a head-on collision with a 0. 300-kg glider that is moving to the left with a speed of 2. 20 m>s. Find the final velocity?

Respuesta :

The final velocity is v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s

Given data,

Mass of glider A (M[tex]_{1}[/tex]) = 0.15 kg

Mass of glider B (M[tex]_{2}[/tex]) = 0.3 kg

Initial velocity of A (u[tex]_{1}[/tex]) = 0.80ms-1

Initial velocity of B ( u[tex]_{2}[/tex]) = -2.2m/s

Momentum and kinetic energy are conserved in elastic collision . So,

M[tex]_{1}[/tex]u[tex]_{1}[/tex]+M[tex]_{2}[/tex]u[tex]_{2}[/tex] = M[tex]_{1}[/tex]v[tex]_{1}[/tex]+M[tex]_{2}[/tex]v[tex]_{2}[/tex]

0.15×0.8+0.3x(-2.2) = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex]

-0.54 = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex]

Again if,

u[tex]_{1}[/tex]+v[tex]_{1}[/tex] =u[tex]_{2}[/tex]+v[tex]_{2}[/tex]

0.8 +v[tex]_{1}[/tex] =-2.2+v[tex]_{2}[/tex]

v[tex]_{1}[/tex]-v[tex]_{2}[/tex] = -3

Solving  for -0.54 = 0.15v[tex]_{1}[/tex]+0.3v[tex]_{2}[/tex] and v[tex]_{1}[/tex]-v[tex]_{2}[/tex] = -3 , we get

v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s

Therefore,The final velocity is v[tex]_{2}[/tex] =-0.2m/s , v[tex]_{1}[/tex] = -3.2 m/s

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