Question

A body of mass m falls from a height h onto a pan (of negligible mass) of a spring balance as shown. The spring also possesses negligible mass and has spring constant k. Just after striking the pan, the body starts socillatory motion in vertical directioin of amplitude A and energy E. Then
(a) A = $\frac{mg}{k}$
(b) A = $\frac{mg}{k}\sqrt{1+\frac{2kh}{mg}}$
(c) E = $mgh+\frac{1}{2}kA^{2}$
(d) E = $mgh+(\frac{2mg}{2k})^{2}$

Answer 1

As the pan is of negligible mass, there is no loss of kinetic energy even though the collision is inelastic. the mechanical energy of the body m in the field generated by the joint action of both the gravity force and the elastic force is conserved i.e. ΔE = 0.during the motion of the body m from the initial to the final (position of maximum compression of the spring) position ΔT = 0, and therefore ΔU = ΔUgr + ΔUsp = 0

If the body m were at rest on the spring, the corresponding position of m will be its equilibrium position and at this Δx (say) due to the body m will be given position the resultant force on the body m will be zero. therefore the equilibrium compression by

Therefore seperation between the equilibrium position and one of the extreme position i.e. the sought amplitude

The mechanical energy of oscillation which is conserved equals E = Uextreme, because at the extreme position kinetic energy becomes zero.

Although the weight of body m is a conservation force, it is not restoring in this problem, hence Uextreme is only concerned with the spring force. therefore

Refer to images below for working

Answer 2

Answer:

Explanation:

As the pan is of negligible mass, with height h, there will be no loss of kinetic energy even though the collision is inelastic. Mechanical energy of the body m in the field is generated by joint action of both the gravity force and the elastic force. The force is conserved i.e. ΔE = 0.

Thus,during the motion of the body m from the initial to the final position ΔT = 0, and therefore ΔU = ΔUgr + ΔUsp = 0

or -mg (h+x) + 1/2kx² = 0

y = mg/k +mg/k √ 1+2kh/mg

At equilibrium mg = ky0

y0 = mg/k

Amplitude A  = y-y0 = mg/k √ 1+2kh/mg

Energy of oscillation = 1/2kA² = mgh + (mg)²/2k