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!
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)
+37157 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 ... 
