Question

$$\begin{gathered} \left(0.10kg+0.20kg\right)v_{1x}=(0.10kg)(3.0m/s) \\ \frac{1}{2}(0.30kg)(0m/s^2)+\frac{1}{2}(3.0N/m){(∆x_2)}^2 \\ =\frac{1}{2}(0.30kg){\left(v_{1x}\right)}^2+\frac{1}{2}(3.0N/m){(0m)}^2 \end{gathered}$$

a. Write a realistic problem for which this is the correct equation(s).

b. Finish the solution of the problem, including a pictorial representation.

Short answer

a) The realistic problem will be "100g ball moves with speed 3.0 m/s collides with 200g ball at rest and move together towards an attached spring 3.0 N/m to a wall. Find the displacement of the spring after the collision of the combination with it"

b) Displacement,$∆x_2=0.18m$

Step-by-step solution

Step 1. Given equation is :0.10 kg+0.20 kgv1x=(0.10 kg)(3.0 m/s)12(0.30 kg)(0 m/s2)+12(3.0 N/m)(∆x2)2 =12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2

We need to find out a realistic problem for which this is the correct equation and have to find out the solution including pictorial representation.

Step 2. Expressing the realistic problem of the given equation 

The realistic problem will be :

"100g ball moves with speed 3.0 m/s collides with 200g ball at rest and move together towards an attached spring 3.0 N/m to a wall. Find the displacement of the spring after the collision of the combination with it"

Step 3. Solving the equations to find the speed of the combination and the displacement.

$$\begin{gathered} Solvingthfirstequation, \\ (0.10kg+0.20kg)v_{1x}=(0.10kg)(3.0m/s) \\ {v}_{1x}=\frac{0.30kgm/s}{0.30kg}=1m/s \\ Solvingsecondequationfor∆x_2, \\ \frac{1}{2}(0.30kg)(0m/s^2)+\frac{1}{2}(3.0N/m){(∆x_2)}^2=\frac{1}{2}(0.30kg){\left(v_{1x}\right)}^2+\frac{1}{2}(3.0N/m){(0m)}^2 \\ \frac{1}{2}(3.0N/m){(∆x_2)}^2=\frac{1}{2}(0.30kg){\left(v_{1x}\right)}^2 \\ {(∆x_2)}^2=0.10{\left(v_{1x}\right)}^2 \\ {(∆x_2)}^2=0.10{(1.0)}^2 \\ ∆x_2=0.18m \end{gathered}$$

Pictorial Representation,