JEE PHYSICS MCQ: LAWS OF MOTION

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Application +4 | -1

Two blocks of equal mass

m

are connected by an
unstretched spring and the system is kept at rest on a
frictionless horizontal surface. A constant force

F

is
applied on one of the blocks pulling it away from the
other as shown in figure.
(a) Find the position of
the centre of mass at time

t.

(b) If the extension of the spring is

x_0

at time

t,

find the displacement of the two
blocks at this instant.

View answer and explanation

Correct option: A

(a)

\frac{F t^2}{4 m};

(b)

x_1=\frac{1}{2}\left(\frac{F t^2}{2 m}-x_0\right), x_2=\frac{1}{2}\left(\frac{F t^2}{2 m}+x_0\right)

Option A ((a)

\frac{F t^2}{4 m};

(b)

x_1=\frac{1}{2}\left(\frac{F t^2}{2 m}-x_0\right), x_2=\frac{1}{2}\left(\frac{F t^2}{2 m}+x_0\right)

) is correct because it accurately calculates the center of mass position and the displacement of each block considering the spring extension.

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