Question

A block having a mass of0.72 kg is given an initial velocity v_A = 1.3 m/s to the right and collides with a spring whose mass is negligible and whose force constant is k = 45 N/m as shown in the figure.

The spring is now mounted vertically on the table, and the mass is dropped downwards, hitting the spring and compressing it. Just before the "collision", the block has a measured velocity of 2.00 m/s downwards. What will be the maximum compression of the spring? (Friction is negligible.)

Answer 1

1) According to the conservation of energy

        ( 1/ 2) m v² = ( 1/ 2) kx2

   maximum compression of the spring is

x = sqrt (m /k) v

                 = sqrt (0.72 kg / 45 N/m)(1.3 m/s)

                 = 0.16 m

2)

According to the conservation of energy

     ( 1/ 2 m v² ) + m g x = ( 1/2) k x²

      (0.5)(0.72 kg ) ( 2 m/s)² + (0.72) ( 9.8 m/s² ) x = ( 45 N/m ) / 2 x²

       22.5 x² - 7.05 x - 1.44 = 0

After solving the quadratic equation we have two value of x

         x = -0.14 m or 0.45 m.

negative value is impossible. so

        x = 0.454 m