Time for block B to stop after collision with ball A

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Homework Statement



A 1-lb ball A is traveling horizontally at 20 ft/s when it strikes a 10-lb block B that is at rest. If the coefficient of restitution between A and B is e = 0.6, and the coefficient of kinetic friction between the plane and the block is µk = 0.4, determine the time for the block B to stop sliding.

The Attempt at a Solution



M1V1=M2V2

(1/32.2)(20) + 0 = (1/32.2)(Va2) + (10/32.2)(Vb2)

(Va2) + 10(Vb2) = 20

e = (Vb2-Va2)/(Va1-Vb1)

0.6 = (Vb2-Va2)/(20-0)

(Vb2 - Va2) = 12

From here I get that

Vb2 = 2.909 ft/s to the right

Va2 = -9.091 ft/s to the left

Block B

T1 + U(1→2) = T2

this where I am stuck ?

any help?
 

Discussion

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So, you need to find the distance B slides. You know the initial speed of B at the beginning of the slide and you know it will slide until it comes to rest.

You can either use energy concepts to get the distance of slide, or you can use Newton's second law to find the deceleration during the slide and then use kinematic equations to find the distance.

If you use energy concepts then you would think about the relationship between the work done by the force of friction and the change in kinetic energy.
 


no, that i need help to determine the time for the block B to stop sliding.
 


Sorry. I misread the question. I would suggest finding the deceleration and using kinematic equations.