Q. In the figure, the ball (A) is released from rest when the spring is at its natural (unstretched) length. For the block (B) of mass (M) to leave contact with the ground at some stage, find: The maximum speed of the block (A) in case, A has the mass found in part (i) which asks about the minimum mass of A.

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The minimum mass of A (m) for B (M) just to lift off is

m = M

The figure with the forces acting on mass A is below.

The equation of motion for mass A is

`F =M*a`

`M*g -k*x =M*a`

`M*g -k*x =M*(d^2x)/(dt^2)`

that can be rearranged in the form

`(d^2x)/dt^2 + (k/M)*x = g`

and by defining `omega = sqrt(k/M)` we have

`(d^2x)/(dt^2) +omega^2*x =g`

with the general solution

`x(t) = X0*cos(omega*t) + C`

by inserting the general solution back into the equation we have

`-omega^2*X0*cos(omega*t) + omega^2*X0*cos(omega*t) +omega^2*C =g`

Hence `C = g/omega^2`

At `t =0` we have  `x(0) =0`

`0 = X0 + g/omega^2`

Hence `X0 =-g/omega^2`

Therefore

`x(t) =-(g/omega^2)*cos(omega*t) +g/omega^2`

The speed of mass A is

`v(t) = dx/dt = ((g*omega)/omega^2)*sin(omega*t)`

and its maximum value is

`v_max = g/omega =g*sqrt(M/k)`

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