Question

A 11.2kg block and a 21.5kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant = 1323N/m). The spring is compressed by 0.144m from its unstrained length and is not attached permanently to either block. What is the speed of the lighter block after the spring is released?

Answer 1

m1 = 11.2 kg
m2 = 21.5 kg
k = 1323 N/m
x = 0.144 m
Let the speed of lighter block be v1 , & of heavier block be v2 !!

Using momentum conservatin,
Initial Momentum = Final Momentum
0 = 11.2 * v1 - 21.5 * v2
v2 = 0.521 v1

Using Energy Conservation,
Initial Spring Potential Energy = Final Kinetic Energy
1/2 * k * x^2 = 1/2 * m1 * v1^2 + 1/2 * m2 * v2^2
1323 * 0.144^2 = 11.2 * v1^2 + 21.5 * (0.521*v1)^2
v1 = 1.27 m/s
Speed of the lighter block, v1 = 1.27 m/s