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Question

The string shown in the figure is passing over a small smooth pulley rigidly attached to trolley A. If the speed of the trolley is constant and equal to vA, speed and acceleration of block B at the instant is vB and aB respectively, which of the following option(s) is/are correct?

A
vB=vA,aB=0
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B
aB=0
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C
vB=35vA
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D
aB=16v2A125
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Solution

The correct option is D aB=16v2A125


From the figure,
l1+l2=constant
(segments of same string)
(yh)+x2+h2=constant
Differentiating w.r.t time t, we get
dydt+xx2+h2dxdt=0
dydt=xx2+h2dxdt
At the instant shown in figure,
xx2+h2=332+42=35
dxdt=vA
dydt=35vA
Since,
vB=(dxdt)2+(dydt)2>dydt
|vB|>35vA
We will now differentiate dydt to obtain acceleration of B.
dydt=xx2+h2.vA
d2ydt2=(x2+h2)1/2.1.dxdtx.12(x2+h2)1/2.2x.dxdt(x2+h2).vA
( vA is constant)
dydt=h2(x2+h2)3/2.dxdt.vA
=v2Ah2(x2+h2)3/2
at the instant shown in figure,
aB=v2A42(32+42)3/2=v2A42(5)3=16125v2A

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