What is the cue ball's momentum after the collision? What is the 5ball'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 5ball, which has a mass of 16 g. After the collision, the 5ball has a speed of 10.03 m/s while the cue ball moves at 1.53 m/s.
A cue ball of mass 23 g hits the 5ball, which has a mass of 16 g. After the collision, the 5ball has a speed of 10.03 m/s while the cue ball moves at 1.53 m/s.
1 Answer
 Cue ball momentum after =
#0.035 kg.m.s^(1)#  5Ball’s KE after =
#0.80 m.s^(1)#  Cue ball momentum before =
#0.20 kg.m.s^(1)#  Cue ball speed before =
# 8.5 m.s^(1)#
Explanation:
Terms:
Subscript
(e.g.
Subscript
Convert mass values to kg:
Cue ball:
5 ball:

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)# 
5Ball’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# 
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 5ball 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)# 
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) #