In this problem, we will consider a head-on collision between Block A (mass = 4.54 kg) and Block B (mass = 1.53 kg). There is friction between the blocks and the ground.

At the instant before the collision, Block A's speed is 7.21 m/s and Block B is at rest. At the instant after the collision, Block A is still traveling in the same direction, but its speed has been reduced to 3.81 m/s.The collision lasts for 0.050 s and both objects remain in contact with the ground as they collide. The coefficient of kinetic friction between Block A and the ground is 0.045 and the coefficient of kinetic friction between Block B and the ground is 0.110.

What is the speed of Block B the instant after the collision?


Really appreciate it if anyone could help. Thank you!

 Feb 26, 2021

Force of friction for ffz = u mg  = ma         ug = a   for A

    initial velocity 7.21

      vf= vi + a t       t is given as .05

       vf = 7.21 - .045(9.81)(.05)

          vf = 7.188     but velocity is only    3.81

              7.188 - 3.81 = 3.378 m/s  component of momentum   given up to B

                    so B should be moving initially at   (disregarding friction presently):


                       Using conservation of momentum for the system:

                           momentum A lost / mass B = velocityB                        3.378 (4.54) / 1.53  =10.023 m/s


Now include the friction component for .05 sec

For B    vf = vi - at

                 = 10.023 - .110(9.81)(.05) = 9.97 m/s   velocity of B after the objects separate              ( not positive about this answer)

 Feb 26, 2021

Anyone else have an answer for this ?    I'd like to know if I was wrong and why....or if I was correct.....I kind of 'punted' a bit ...  cheeky

 Feb 27, 2021

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