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) Draw a Free Body Diagram

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

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

Explanation / Answer

Mass of chunk m = 800 g = 0.8 kg

mass of table M = 1600 g = 1.6 kg

Initial speed of chunk u = 1.65 m/s

Initial speed of table U = 0

Speed of the chunk and table system after collision v = ?

From law of conservation od momentum , mu + MU = (m+ M) v

From this v = [mu +MU ]/(m+M)

                  = [(0.8x1.65) +0]/(0.8+1.6)

                  = 1.32 / 2.4

                   = 0.55 m/s

After adding the mass of the chunk the spring compresses an additionally by 0.06 m

We know weight of the chunk = Restoring force of the spring

                                   mg = kx

               0.8kg x9.8m/s 2 = k(0.06m)

     Spring constant k = (0.8x9.8) /0.06

                                = 7.84 / 0.06

                                = 130.66 N/m

                                ~ 130 N/m

We know kinetic energy of the chunk(clay)+table system after colliision = potential energy of the springs

   (1/2) (m+M) v 2 + (m+M) gA= (1/2) kA 2

         (m+M) v 2 +2(m+M) gA = kA 2

   2.4 x 0.55 2 + 47.04 A = 130 A 2

130 A 2 -47.04 A -0.726 = 0

Solve this you get A = 0.096 m                  

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