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A small smooth sphere A of mass m moving on a smooth horizontal table with velocity u...
A small smooth sphere A of mass m moving on a smooth horizontal table with velocity u collides directly with another small smooth sphere B, of equal size and mass 2m, which is at rest on the table.The coefficient of restitution is e.
Show that the velocity acquired by B is (1+e)u/3.
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You need to test if the velocity aquired by B is `(1+e)u/3` , hence, you should use the equation of the velocity of sphere B after a one dimensional inelastic collision, such that:
v_B = (C_Rm_A*(u_A-u_B) + m_A*u_A + m_B*u_B)/(m_A+m_B)
Since u_A = u, u_B = 0 represents the velocities of spheres before collision, yields:
`v_B = (e*m(u - 0) + m*u + 2m*0)/(m + 2m)`
`v_B = (e*m*u + m*u)/(3m)`
Factoring out m*u to numerator yields:
`v_B = (m*u(e + 1))/(3m)`
Reducing duplicate factors yields:
`v_B = u*(e + 1)/3`
Hence, evaluating the velocity acquired by B after collision, yields `v_B = u*(e + 1)/3.`
Posted by sciencesolve on July 4, 2013 at 6:12 PM (Answer #1)
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