Engineering Mechanics - PKRB: Impulse and Momentum - Discussion

Discussion :: PKRB: Impulse and Momentum - General Questions (Q.No.1)


The uniform rod AB has a weight of 3 lb and is released from rest without rotating from the position shown. As it falls, the end A strikes a hook S, which provides a permanent connection. Determine the speed at which the other end B strikes the wall at C.

[A]. vB4 = 14.74 ft/s
[B]. vB4 = 20.3 ft/s
[C]. vB4 = 22.0 ft/s
[D]. vB4 = 12.69 ft/s

Answer: Option B


No answer description available for this question.

Supratik said: (Jun 26, 2014)  
How this possible ? can you please explain me.

Nishant said: (Apr 16, 2016)  
Can anybody solve this?

Adarsh said: (Jun 20, 2016)  
As per my calculation, it's 14.74.

So the answer is option A.

Dks said: (Oct 2, 2018)  
A is correct by considering 3.5ft.

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