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# physics question

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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

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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

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 ... 