a) Use the worked example above to help you solve this problem. Two billiard bal

Question

a) Use the worked example above to help you solve this problem. Two billiard balls of identical mass move toward each other as shown in the figure. Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are +28.8 cm/s and -19.2 cm/s, what is the velocity of each ball after the collision? Assume friction and rotation are unimportant.

v1f =
v2f =

b) Find the final velocity of the two balls if the ball with velocity v2i = -19.2 cm/s has a mass equal to half that of the ball with initial velocity v1i = +28.8 cm/s.

v1f =
v2f =

Explanation / Answer

Given  mass of two balls   m1  = m2  = m initial velocities of two balls   u1  = 28.8 cm/s= 0.28 m/s                                              u2  = - 19.2  cm /s  = - 0.192  m/s   From law of conservation of  momentum          m1 u1  + m2 u2  = m1v1 + m2v2           m1 ( u1-  v1 )  = m2  ( v2 - u2 )  ...................(1)  From law of conservation of  energy           ( 1/2)m1 u12 + (1/2)  m2 u22  = (1/2) m2 v12  + ( 1/2) m2 v22                   m1 ( u12 - v12  )  =  m2 ( v22 - u22 ) ...........(2)  divide equation (2)  by  (1)                  u1  + v1  = u2 + v2                        v1  =    u2 + v2  - u1 ..........(3)                               =   -0.192 + v2  - 0.288                               =   v2  - 0.48  ........(4)        substitute  v1 value in equation (1)                u1  -  v2  + 0.48  =  v2  - u2                  u1 + u2 + 0.48  = 2 v2                            v2  =  0.288-0.192+0.48  / 2                                =   0.288 m/s  substitute  v2 value in equation  (4)                         v1  =  0.288 - 0.48                               =  -0.192  m/s   _______________________________________________________ Mass of first ball is   m1  = m   mass of second ball is   m2  = m /2 initial velocities of two balls   u1  = -19.2  cm/s=-  0.192m/s                                              u2  = 28.8 cm /s  = 0.288  m/s    substitute  v1  value  in equation  (1)         u1  -  v2  + 0.48  =  ( v2  - u2 ) /2                   2u1 + u2 + 2*0.48  = 3 v2                            v2  = 0.448  m/s      substitute   v2  value  in equation  (4)                         v1  = 0.448-0.48                             =  -0.032  m/s                                                   _______________________________________________________ Mass of first ball is   m1  = m   mass of second ball is   m2  = m /2 initial velocities of two balls   u1  = -19.2  cm/s=-  0.192m/s                                              u2  = 28.8 cm /s  = 0.288  m/s  initial velocities of two balls   u1  = -19.2  cm/s=-  0.192m/s                                              u2  = 28.8 cm /s  = 0.288  m/s    substitute  v1  value  in equation  (1)         u1  -  v2  + 0.48  =  ( v2  - u2 ) /2                   2u1 + u2 + 2*0.48  = 3 v2                            v2  = 0.448  m/s      substitute   v2  value  in equation  (4)                         v1  = 0.448-0.48                             =  -0.032  m/s                                   2u1 + u2 + 2*0.48  = 3 v2                            v2  = 0.448  m/s      substitute   v2  value  in equation  (4)                         v1  = 0.448-0.48                             =  -0.032  m/s                                       

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.