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Solve the foregoing problem for the case of the pan having a mass `M`. Find the oscillation amplitude in this case. |
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Answer» Unlike the previous `(4.40)` problem the kinetic energy of body `m` decreases due to the perfelcty inelastic collision with the pan. Obviously the body `m` comes to strike the pan with velocity `v_(0)=sqrt(2gh)`. If `v` be the common velocity of the `'` body `m+` pena `"` system due to the collision thrn from the conservation of linear momentum `m v_(0)=(M+m)v` or `v=(m v_(0))/((M+m))=(msqrt(2gh))/((M+m)) ...(1)` At the moment the body `m` strikes the pan, the spring is compressed due to weight of the pan by the amount `Mg//k`. If `l` be the further compression fo the spring due to the velocity acquired by the `"` pan `-` body `m"` system, then from the conservatiion of mechanical energy of the said system in the field generated by the joint action of boty the gravity and spring forces `(1)/(2)(M+m)v^(2)+(M+m)gl=(1)/(2)k((Mg)/(k)+l)^(2)-(1)/(2)k((Mg)/(k))^(2)` or, `(1)/(2)(M+m)(m^(2)2gh)/((M+m))+(M+m)gl=(1)/(2)k((Mg)/(k))^(2)+(1)/(2)kl^(2)-Mgl-(1)/(2)k ((Mg)/(k))^(2)` (Using 1) or, `(1)/(2)kl^(2)-mgl-(m^(2)gh)/((m+M))=0` Thus` l=(mg+-sqrt(m^(2)g^(2)+(2kghm^(2))/(M+m)))/(k)` As minus sign is not acceptable `l=(mg)/(k)+(1)/(k)sqrt(m^(2)g^(2)+(2km^(2)gh)/((M+m)))` If the oscillating " pan `+` body `m"` system were at rest it correspong to their equilibrium position i.e. the spring were compressed by `((M+m)g)/(k)` therefore the amplitude of oscillation `a=l-(mg)/(k)=(mg)/(k)sqrt(1+(2hk)/(mg))` The mechanical energy of oscillation which is only conserved with the restoring forces becomes `E=U_("extreme")=(1)/(2)ka^(2)`( Because spring force is the only restoring force not the weight of the body ) Alternately `E=T_("mean")=(1)/(2)(M+m)a^(2)omega^(2)` thus `E=(1)/(2)(M+m)a^(2)((k)/(M+m))=(1)/(2)ka^(2)` |
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