Answer
a) $4.62\;\rm cm$
b) $1.33\;\rm m/s$, $5.33\;\rm m/s$
Work Step by Step
Let's assume that the two carts are rolling without friction and that the spring is massless and obeying Hooke's law; which means that the system [the two carts+the spring] is an isolated system.
So the momentum and energy of the system are conserved.
We have three stages here,
1) Before the collision. In this stage, the 2-kg cart is moving toward the stationary 1-kg cart.
2) When the spring is compressed at maximum compression during the collision. In this stage, the two carts are moving as one unit. Thus, the two carts are having the same velocities.
3) After the collision. In this stage, each cart is moving on its own.
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a) To find the maximum compression of the spring, we need to use the conservation of energy for the first stage and the second stage.
$$E_1=E_2$$
$$K_1+\overbrace{U_{s1}}^{=0}=K_2+U_{s2}$$
$$\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2v_{i2}^2=\frac{1}{2}m_1v_{f1}^2+\frac{1}{2}m_2v_{f2}^2+\frac{1}{2}kx^2$$
The second cart was initially at rest, and the two carts when the spring in at its maximum compression is moving as one unit.
So,
$$\frac{1}{2}m_1v_{i1}^2+\frac{1}{2}m_2(0)^2=\frac{1}{2}(m_1+m_2)v_{f1}^2 +\frac{1}{2}kx^2$$
$$\color{red}{\bf\not}\frac{1}{2}m_1v_{i1}^2 =\color{red}{\bf\not}\frac{1}{2}(m_1+m_2)v_{f1}^2 +\color{red}{\bf\not}\frac{1}{2}kx^2$$
$$ m_1v_{i1}^2 - (m_1+m_2)v_{f1}^2 = kx^2$$
Thus,
$$x=\sqrt{\dfrac{m_1v_{i1}^2 - (m_1+m_2)v_{f1}^2}{k}}\tag 1$$
Now to find the final speed of the two carts as one unit, we need to apply the conservation of momentum during the first and the second stages.
$$p_{ix}=p_{fx}$$
$$m_1v_{i1}=(m_1+m_2)v_{f1}$$
Thus,
$$v_{f1}=\dfrac{m_1v_{i1}}{(m_1+m_2)}\tag 2$$
Plugging into (1);
$$x=\sqrt{\dfrac{m_1v_{i1}^2 - (m_1+m_2)\left(\dfrac{m_1v_{i1}}{(m_1+m_2)}\right)^2}{k}} $$
$$x=\sqrt{\dfrac{m_1v_{i1}^2 - \dfrac{m_1^2v_{i1}^2}{(m_1+m_2)} }{k}} $$
Plugging the known;
$$x=\sqrt{\dfrac{[2\times 4^2] - \dfrac{[2^2\times 4^2]}{(2+1)} }{5000}}=\bf 0.04618\;\rm m $$
Therefore, the maximum compression is
$$x=\color{red}{\bf 4.62}\;\rm cm$$
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b) Now we need to apply the conservation of energy and the conservation of the momentum for the second and the third stages.
$$p_{ix}=p_{fx}$$
where $i\rightarrow$ for the second atge and $f\rightarrow$ for the third stage.
$$(m_1+m_2)v_{f1}=m_1v_{1}'+m_2v_2'$$
Plugging from (2);
$$(m_1+m_2)\dfrac{m_1v_{i1}}{(m_1+m_2)}=m_1v_{1}'+m_2v_2'$$
$$ m_1v_{i1} =m_1v_{1}'+m_2v_2' $$
Plugging the known;
$$ 2\times 4 =2v_{1}'+v_2' $$
Thus,
$$v_2' =8-2v_{1}'\tag 3$$
And
$$E_2=E_3$$
$$K_2+U_{s2}=K_3+U_{s3}$$
$$\frac{1}{2}(m_1+m_2)v_{f1}^2 +\frac{1}{2}kx^2=\frac{1}{2} m_1{v_{1}'}^2+\frac{1}{2}m_2{v_{2}'}^2+\frac{1}{2}k(0)^2$$
Plugging from (2);
$$\color{red}{\bf\not}\frac{1}{2}(m_1+m_2)\left[\dfrac{m_1v_{i1}}{(m_1+m_2)}\right]^2 +\color{red}{\bf\not}\frac{1}{2}kx^2=\color{red}{\bf\not}\frac{1}{2} m_1{v_{1}'}^2+\color{red}{\bf\not}\frac{1}{2}m_2{v_{2}'}^2$$
$$ \dfrac{m_1^2v_{i1}^2}{ m_1+m_2 } + kx^2= m_1{v_{1}'}^2+ m_2{v_{2}'}^2$$
Plugging the known;
$$ \dfrac{2^2\times 4^2}{ 2+1} + (5000\times 0.04618^2)= 2{v_{1}'}^2+ {v_{2}'}^2$$
$$ 32= 2{v_1^2}'+ {v_2^2}'$$
Plugging from (3);
$$ 32= 2{v_1^2}'+ (8-2v_{1}')^2$$
$$ 32= 2{v_1^2}'+ 64-32{v_{1}'}+4{v_{1}'}^2$$
Thus,
$$ 6{v_1^2}'-32{v_{1}'}+ 32=0$$
Hence,
$${v_{1}'}=1.33\;{\rm m/s}$$
Or,
$${v_{1}'}=4\;{\rm m/s}$$
Plugging these two results into (3);
Hence,
$${v_{2}'}=5.33\;{\rm m/s}$$
Or,
$${v_{2}'}=0\;{\rm m/s}$$
Now we have two results,
- If ${v_{1}'}=1.33\;{\rm m/s}$, then ${v_{2}'}=5.33\;{\rm m/s}$
- If ${v_{1}'}=4\;{\rm m/s}$, then ${v_{2}'}=0\;{\rm m/s}$
It seems that the second result is not suitable since it gave the first car the same speed as if there is no collision at all (as if the cart is moving forever without hitting the second cart). Also, it is impossible that the speed of the second cart will be zero suddenly after the collision.
Therefore, the two cart will be moving on the same direction as the first cart was heading before the collision, and their speeds are as follows;
$${v_{1}'}=\color{red}{\bf 1.33}\;{\rm m/s}$$
$${v_{2}'}=\color{red}{\bf 5.33}\;{\rm m/s}$$
