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Standard XII

Physics

Conservation of Momentum

A pulley fixe...

Question

A pulley fixed to the ceiling of a lift carries a string whose ends are tied to blocks of mass 2 kg and 1 kg. The lift is having a downward acceleration 2 $m / s^{2}$ .
Calculate the acceleration of the block of mass m = 1 kg relative the lift.

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8/3

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2/3

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14/3

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32/3

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Solution

The correct option is A $-$ 8/3
$\begin{matrix} 1 g - T = 1 a_{1} . . . . . . . . (1) 2 g - T = 2 a_{2} . . . . . . . . . (2) \frac{a_{1} + a_{2}}{2} = 2 \Rightarrow a_{1} + a_{2} = 4......... (3) \Rightarrow g - T = a_{1} g - \frac{T}{2} = a_{2} \Rightarrow 2 g - \frac{3 T}{2} = 4 \Rightarrow 20 - 4 = \frac{3 T}{2} \Rightarrow T = \frac{16 \times 2}{3} = \frac{32}{3} 10 - \frac{32}{3} = a_{1} \Rightarrow - \frac{2}{3} = a_{1} \Rightarrow w i t h r e s p e c t t o p u l l y a = \frac{- 2}{3} - 2 = \frac{- 8}{3} H e n c e, o p t i o n A i s c o r r e c t a n s w e r . \end{matrix}$

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A pulley fixed to the ceiling of a lift carries a string whose ends are tied to blocks of mass 2 kg and 1 kg. The lift is having a downward acceleration 2 m/s2. Calculate the acceleration of the block of mass m = 1 kg relative the lift.

The correct option is A −8/31g−T=1a1........(1)2g−T=2a2.........(2)a1+a22=2⇒a1+a2=4.........(3)⇒g−T=a1g−T2=a2⇒2g−3T2=4⇒20−4=3T2⇒T=16×23=32310−323=a1⇒−23=a1⇒withrespecttopullya=−23−2=−83Hence,optionAiscorrectanswer.

A pulley fixed to the ceiling of a lift carries a string whose ends are tied to blocks of mass 2 kg and 1 kg. The lift is having a downward acceleration 2 $m / s^{2}$ .
Calculate the acceleration of the block of mass m = 1 kg relative the lift.