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.
1
Expert's answer
2020-11-03T10:31:46-0500

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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