Question

A spring of negligible mass and force constant k = 410 N/m is hung vertically, and a 0.210 kg pan is suspended from its lower end. A butcher drops a 2.4 kgsteak onto the pan from a height of 0.50 m . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM.

a) What is the speed of the pan and steak immediately after the collision?

b) What is the amplitude of the subsequent motion?

c) What is the period of that motion?

Answer 1

moments are conserved, when the steak hits the pan:
velocity v of the steak:
v = ?(2gh) = ?(2*9.81*0.25) = 2.215 m/s
velocity pan + steak is
v = m1v1/(m1+m2) = 2.5*2.215/(2.7) = 2.050 m/s
---------
2)
old equilibrium point: x = F/k = 0.2*9.81/420 = 0.00467 m below the end of the relaxed spring
new equilibrium point: x = F/k = 2.7*9.81/420 = 0.063 m below the end of the relaxed spring.

let the motion start at the new equilibrium point. In fact it starts 6 cm above this point.
---------
3).
initial kinetic energy of pan + steak = spring energy
1/2 mv^2 = 1/2 kA^2 with A = amplitude
2.7* 2.05^2 = 420*A^2
A = 0.1643 cm ---> amplitude
------
4)
T = 2pi?(m/k) = 2pi?(2.7/420) = 0.503 s
-----
5
yes, at the highest point of the movement