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Question

As shown in the figure, two masses 6 kg and 4 kg respectively are tied to the ends of a string which passes over a light frictionless pulley. Initially, the masses are at rest and are at the same horizontal level. Now, they are released. Find the distance travelled by their center of mass in 10 s.

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Solution

The correct option is **B** 20 m

Assume acceleration of the system is a, as shown.

F.B.D of 6 kg block:

6g−T=6a

T=6(g−a) ...(1)

F.B.D of 4 kg block:

T−4g=4a

T=4(g+a) ...(II)

From eq. (I) and (II)

4(g+a)=6(g−a)

6g−4g=4a+6a

10a=2g

or a=2 m/s2

Distance travelled by 6 kg mass in 10 s is (taking downward +ve)

S=ut+12 at2

Here u=0 (initially at rest)

⇒S=12×2×102=100 m. {Downwards}

Similarly, distance travelled by 4 kg mass block is 100 m {upwards}

So, the distance travelled by COM is

S′=6×(100)+4(−100)6+4

S′=20010=20 m {Downwards}

Hence, the distance travelled by the COM of the blocks in 10 s is 20 m. {Downwards}

Assume acceleration of the system is a, as shown.

F.B.D of 6 kg block:

6g−T=6a

T=6(g−a) ...(1)

F.B.D of 4 kg block:

T−4g=4a

T=4(g+a) ...(II)

From eq. (I) and (II)

4(g+a)=6(g−a)

6g−4g=4a+6a

10a=2g

or a=2 m/s2

Distance travelled by 6 kg mass in 10 s is (taking downward +ve)

S=ut+12 at2

Here u=0 (initially at rest)

⇒S=12×2×102=100 m. {Downwards}

Similarly, distance travelled by 4 kg mass block is 100 m {upwards}

So, the distance travelled by COM is

S′=6×(100)+4(−100)6+4

S′=20010=20 m {Downwards}

Hence, the distance travelled by the COM of the blocks in 10 s is 20 m. {Downwards}

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