# Q. A block of mass `m` is attached to a pulley disc of equal mass `m` ,radius `r` by means of a slack string as shown.The pulley is hinged about its centre on a horizontal table and the block is projected with an initial velocity of 5 m/s.Its velocity when the string becomes taut will be:- A) 3 m/s B) 2.5 m/s C) 5/3 m/s D) 10/3 m/s

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The theorem of momentum variation applied to mass on the table says:

`F =(Delta(P))/(Delta(t)) = (m(v_2-v_1))/t`

where `v_2` is the final speed and `v_1` the initial speed of the mass `m`

For the pulley the theorem of angular momentum variation says:

`M =(Delta(L))/(Delta(t)) =(I*omega_2)/t`

where `M=(-F)*r`

(here the minus sign shows the force on the pulley is opposing to the force on the mass on the table)

Therefore we have two equations:

`(-F)*r =(I*omega_2)/t`

`F = (m(v_2-v_1))/t`

and we obtain

`m(v_1-v_2)*r =I*omega_2`  (1)

Now we can  write for the angular acceleration of the pulley and its momentum of inertia (the pulley is considered a solid cylinder)

`omega_2 =v_2/r`  and `I =(m*r^2)/2`

Replacing these two relations in (1) we get

`m(v_1-v_2)*r =(m*r^2)/2*v_2/r`

`2(v_1-v_2) =v_2`

`2v_1 =3v_2`

`v_2 =(2/3)*v_1 =2/3*5 =10/3 =3.33 m/s`

The velocity of the mass on the table when the string is taut will be `v_2 =10/3 m/s`.

The correct answer is D) 10/3

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