What is the cue ball's momentum after the collision? What is the 5-ball's kinetic energy after the collision? What was the cue ball's momentum before the collision? What was the speed of the cue ball before the collision?

A cue ball of mass 23 g hits the 5-ball, which has a mass of 16 g. After the collision, the 5-ball has a speed of 10.03 m/s while the cue ball moves at 1.53 m/s.

1 Answer
Dec 20, 2015

Answer:

  1. Cue ball momentum after = #0.035 kg.m.s^(-1)#
  2. 5-Ball’s KE after = #0.80 m.s^(-1)#
  3. Cue ball momentum before = #0.20 kg.m.s^(-1)#
  4. Cue ball speed before = # 8.5 m.s^(-1)#

Explanation:

Terms:
#u# is an initial velocity
#v# is a final velocity
#m# is a mass
#p# is a momentum
Subscript #c# denotes a property of the cue ball
(e.g. #m_c# is the mass of the cue ball.)
Subscript #5# denotes a property of the 5-ball

Convert mass values to kg:
Cue ball: #m_c= 23/1000 = 0.023 g#
5 ball: #m_5=16/1000 = 0.016 g#

  1. Cue ball’s momentum after the collision
    #p_(c2) = m_c × v_c = 0.023 × 1.53 = 0.03519 #
    #⇒ p_(c2) = 0.035 kg.m.s^(-1)#

  2. 5-Ball’s kinetic energy after the collision
    #E_k=½ m v^2 = ½ × 0.016 × 10.03^2 = 0.8048072 #
    #⇒ E_k = 0.80 J#

  3. Use conservation of momentum to calculate the momentum of the cue ball before the collision
    Total momentum before = total momentum after
    Note that I am assuming that the 5-ball was stationary before the collision (the question does not state this). I am also assuming that the two balls move off in the same direction.
    #p_(c1) + 0 = p_(c2) + m_5 × v_5#
    #p_(c1)# is the cue ball’s momentum before the collision.
    #p_(c1) = 0.03519 + 0.016 × 10.03 = 0.19567#
    # ⇒ p_(c1) = 0.20 kg.m.s^(-1)#

  4. Calculate the speed of the cue ball before the collision
    #p_(c1) = m_c × u_c ⇒ u_c = p_(c1) / m_c = 0.19567 / 0.023 = 8.5073913 #
    # ⇒ u_(c) = 8.5 m.s^(-1) #