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Standard XII

Physics

COE in SHM

An elastic sp...

Question

An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12 J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be

2 m/s

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4 m/s

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1 m/s

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8 m/s

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Solution

The correct option is A 2 m/s
Using momentum conservation
$m_{A} . v_{A} = m_{B} v_{B}$
$\Rightarrow v_{A} = 2 v_{B}$
Now total $K . E = P . E$ of spring

$\frac{1}{2} m_{A} v_{A}^{2} + \frac{1}{2} m_{B} {(V_{B})}_{B}^{2} = 12$

$\frac{1}{2} m_{A} \times 4 V_{B}^{2} + \frac{1}{2} m_{B} (V_{B})^{2} = 12$

$3 {(v_{B})}_{B}^{2} = 12$
$v_{B} = 2 m / s$

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An elastic spring is compressed between two blocks of masses 1 kg and 2 kg resting on a smooth horizontal table as shown. If the spring has 12 J of energy and suddenly released, the velocity with which the larger block of 2 kg moves will be

The correct option is A 2 m/sUsing momentum conservationmA.vA=mBvB⇒vA=2vBNow total K.E=P.E of spring12mAv2A+12mB(VB)2=1212mA×4V2B+12mB(VB)2=123(vB)2=12vB=2m/s