A 100g cart is placed against a spring with a force constant of 20,000 N/m. The spring is compressed 2.5cm. A second cart with a mass of 200g is...

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Denote the mass of the first cart as `m_1,` the mass of the second cart as `m_2,` the spring's force constant as `k_s` and the coefficient of friction as `k_f.` Denote the distance of spring compression as `x.`

The friction force will be `0.25*m_1*g approx 0.25N.` The starting spring force is `20000 N/m*0.025 m = 500N.` Therefore we can neglect the friction force while spring is releasing. Also we can neglect the position change while spring is releasing...

Hello!

Denote the mass of the first cart as `m_1,` the mass of the second cart as `m_2,` the spring's force constant as `k_s` and the coefficient of friction as `k_f.` Denote the distance of spring compression as `x.`

The friction force will be `0.25*m_1*g approx 0.25N.` The starting spring force is `20000 N/m*0.025 m = 500N.` Therefore we can neglect the friction force while spring is releasing. Also we can neglect the position change while spring is releasing (2.5cm vs. 5m).

Determine the speed V_0 of the first cart after the spring is released completely. Use the law of energy conservation: the elastic potential energy is `(1/2)*k_s*x^2` transforms into kinetic energy of the first car, `(1/2)*m_1*V_0^2.`

So `V_0=x*sqrt(k_s/m_1) = 0.025*sqrt(200,000) approx 11.2 (m/s).`

Then the first cart decelerates uniformly due to the friction force `k_f*m_1*g.` It gives the negative acceleration of `a=k_f*g approx 2.45 (m/s^2)` (Newton's Second law).

The speed is  `V(t)=V_0-a*t,` the position is  `L(t)=V_0*t-a*t^2/2.`

Let's find the (smallest) time `t_1` when `L(t_1)` becomes 5m (and collision happens):

`2.45t_1^2-22.4t_1+10=0,`

`t_1 = (22.4-sqrt(22.4^2-4*2.45*10))/4.9 approx 0.47 (s).`

The speed before collision will be `V_1=V_0-a*t_1 approx 11.2-2.45*0.47 approx 10 (m/s).`

Now for the collision. Elastic collision means no bounce, both bodies move as a whole. Then the law of impulse conservation gives that the final speed `V_2` satisfies

`m_1*V_1 = (m_1+m_2)*V_2,` or

`V_2 = (m_1/(m_1+m_2))*V_1 = V_1/3 approx 3.33 (m/s).` This is the answer.