Question

A spring with a force constant of 89 N/m is compressed 8.7 cm and placed between two stationary dynamics carts of mass 1.0 kg and 1.5 kg. If friction is negligible, determine the final speed of the more massive cart when the spring is released?

Answer 1

0,60 m/s

Explanation:

The total elastic energy of the system is `E =1/2 k x^2 = 1/2 89 (N/m) 8,7^2 10^(-4) m^2= 0,674 J`
The elastic energy will be the kinetic energy of the system after the spring is released. `1/2 m_1V_1^2 + 1/2 m_2V_2^2= 0,674 J`
The momentum of the two carts will be the same but in opposite direction, so the smaller cart will have 1,5 the velocity of bigger one `V_1 =1,5 V_2`.
So
`1/2 1kg (1,5 V_2)^2 +1/2 1,5kg V_2^2 = 1,125 kg V_2^2 + 0,75kg V_2^2 = 1,875kg V_2^2=E` `V_2=sqrt(E/(1,875kg))= sqrt((0,674J)/(1,875 kg))= 0,60 m/s`