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(A) The weight of the disc is balanced by the average force exerted by the stones on the disc.When a stone strikes the disc and stops (or rebounds),it

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$1.2$

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(A) The weight of the disc is balanced by the average force exerted by the stones on the disc.When a stone strikes the disc and stops (or rebounds),it exerts an impulsive force.Assuming the stones strike and come to rest relative to the disc (or transfer their full momentum),the force $F$ exerted by $n$ stones per second is given by the rate of change of momentum:$F = n \cdot \Delta p = n \cdot m \cdot v$Given $n = 40\,s^{-1}$,$m = 0.05\,kg$,and $v = 6\,m/s$:$F = 40 	imes 0.05 	imes 6 = 12\,N$For the disc to be held in equilibrium,the upward force must equal the weight of the disc:$F = Mg$$12 = M 	imes 10$$M = \frac{12}{10} = 1.2\,kg$

$A$ solid disc of mass $M$ is held horizontally in the air by throwing $40$ stones per second vertically upwards to strike the disc,each with a velocity of $6\,m/s$. If the mass of each stone is $0.05\,kg$,what is the mass of the disc in $kg$? (Take $g = 10\,m/s^2$)

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