Monday, October 18, 2010

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 (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.

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