Question #142029

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. a croquet player strikes the green ball and it gains an initial velocity of 2.26 m/s. It then strikes the blue ball which is initially at rest . After the collision the green ball has a velocity of 1.14 m/s is the same direction. If the ball roll on a frictionless surface and the the collision is head on what is the final velocity of the blue ball.

Expert's answer

We can find the final velocity of the blue ball after the collision from the Law of Conservation of Momentum:

mgvg,i+mbvb,i=mgvg,f+mbvb,f,m_gv_{g,i}+m_bv_{b,i}=m_gv_{g,f}+m_bv_{b,f},

here, mg=0.525 kgm_g=0.525\ kg is the mass of the green ball, mb=0.482 kgm_b=0.482\ kg is the mass of the blue ball, vg,i=2.26 msv_{g,i}=2.26\ \dfrac{m}{s} is the initial velocity of the green ball before the collision, vb,i=0 msv_{b,i}=0\ \dfrac{m}{s} is the initial velocity of the blue ball before the collision, vg,f=1.14 msv_{g,f}=1.14\ \dfrac{m}{s} is the final velocity of the green ball after the collision, vb,fv_{b,f} is the final velocity of the blue ball after the collision.

Then, from this equation we can calculate vb,fv_{b,f}:


vb,f=mg(vg,ivg,f)mb,v_{b,f}=\dfrac{m_g(v_{g,i}-v_{g,f})}{m_b},vb,f=0.525 kg(2.26 ms1.14 ms)0.482 kg=1.22 ms.v_{b,f}=\dfrac{0.525\ kg\cdot(2.26\ \dfrac{m}{s}-1.14\ \dfrac{m}{s})}{0.482\ kg}=1.22\ \dfrac{m}{s}.

Answer:

vb,f=1.22 ms.v_{b,f}=1.22\ \dfrac{m}{s}.


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