Maximum speed of 3 kg block in the subsequent motion will be n × v 0 find the value of n.

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PhysicsSHMEnergyMedium2 minQB

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Passage / Comprehension

2 kg and 3 kg blocks are placed on a smooth horizontal surface and connected by spring which is unstretched initially. The blocks are imparted velocities as shown in the figure.

Maximum speed of 3 kg block in the subsequent motion will be $n \times v_{0}$ find the value of n.

Answer:

3.00

Solution:

The centre-of-mass velocity is

$v_{C} = \frac{( 2 ) ( - 4 v _{0} ) + ( 3 ) ( v _{0} )}{5} = - v_{0}$

Initially the 3 kg block has speed relative to the centre of mass

$v_{3, CM} = v_{0} - (- v_{0}) = 2 v_{0}$

Its maximum speed in the ground frame occurs when its relative motion points left.

$∣ v_{3, m a x} ∣ = ∣ - v_{0} - 2 v_{0} ∣ = 3 v_{0}$

$n = 3$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: QB

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