Please Help!! In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m). a) Determine the velocity of each cart after the collision. b) Determine the maximum compression of the spring.

initial momentum 0.6kg * -5m/s + 0.8kg * 2m/s = -1.4 kg·m/s

final momentum = -1.4 kg·m/s = 0.6kg * u + 0.8kg * v

For an elastic head-on collision, we know that the

relative velocity of approach = relative velocity of separation, or

7m/s = u - v, or

u = v + 7

Plug this into momentum eqn:

-1.4 = 0.6(v+7) + 0.8v = 0.6v + 4.2 + 0.8v

-5.6 = 1.4v

v = -4 m/s ← 0.8kg cart, now moving West

u = v + 7 = 3 m/s ← 0.6kg cart, now moving East

initial Ek = ½ * 0.6kg * (5m/s)² + ½ * 0.8kg * (2m/s)² = 9.1 J

This gets converted into spring energy,

U = 9.1 J = ½kx² = ½ * 1200N/m * x²

x = 0.123 m = 12.3 cm

i wanna say it’s between b and d, leaning more towards b.

i believe the answer is a hope it and is correct