Question

A sphere 'A' is released from rest in the position shown and strikes the block 'B' which is at rest. If $e = 0.75$ between 'A' and 'B' and $μ _{k} = 0.5$ between 'B' and support, determine :

( i ) The velocity of 'A' just after the impact
( ii ) The maximum displacement of 'B' after the impact...
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Answer 1

♊♊YOUR ANSWERS♍♍

✍(i) √(g/12) (ii) 49/48

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EXPLANATION —

↪Actually welcome to the concept Rigid Body Dynamics

(i)

Using conservation of momentum. and definition of e we get

m1v = m1v1+ m2v2 i.e 2v = 2v1 + 4v2

0.75 = (v2- v1) /v. i.e v2 = 0.75v+ v1

Solving

⇒v1 = - v/6

⇒v1 = v/6 = √(2×1.5×g)/6 = √(g/12)

(ii)

Solving above we get v2 = √(49g/48)

all the energy is lost in friction so acc. to work energy theorem.

1/2 m v² = μmgs

⇒s = v²/μg

⇒s = 49g/(2× 48× 0.5 ×g) = 49/48

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