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

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

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