A 800 g chunk of modeling clay is held above a table and dropped so that it hits

Question

A 800 g chunk of modeling clay is held above a table and dropped so that it hits the table with a speed of 1.65 m/s. The table (m = 1600 g) is supported on 4 springs. The clay makes an inelastic collision with the table and the table and the clay oscillate up and down. After a long time the table comes to rest 0.06 m below its original position.

(a) What is the effective spring constant of all four springs taken all together?

(b) What maximum amplitude that the platform oscillates with when it starts bouncing?

Explanation / Answer

part a:

let effective spring constant be k.

total weight supported by the springs=(1600+800)*0.001*9.8=23.52 N

then k*compression=23.52

==>k*0.06=23.52

==>k=392 N/m

part b:

as there is no external force, total mechanical energy of the system will remain constant

assume that initial position of the table top is zero potential energy reference point.

hence initial total mechanical energy=initial kinetic energy of the clay just before it hits the table=0.5*mass*speed^2=0.5*0.8*1.65^2=1.089 J

when it has reached its maximum amplitude, its speed will be zero.

spring's compression will be k*A where A=amplitude

total final energy=potential energy of clay+potential energy of table+potential energy of spring

=0.8*9.8*(-A)+1.6*9.8*(-A)+0.5*392*A^2

=196*A^2-23.52*A

equating initial and final energy values

196*A^2-23.52*A=1.089

==>A=0.1556 m

hence as equilibrium position lies 0.06 m below original position, amplitude=0.1556-0.06=0.0956 m

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