Question

The air-track carts in FIGURE CP11.83 are sliding to the right at 1.0 m/s. The spring between them has a spring constant of 120 N/m and is compressed 4.0 cm. The carts slide past a flame that burns through the string holding them together. Afterward, what are the speed and direction of each cart?

Short answer

Speed of first cart is 1.4 m/s towards Left and speed of second cart is 1.8 m/s towards right.

Step-by-step solution

Step 1. Given Information is Mass of the object 1, m1 = 100 gMass of the object 2, m2 = 300 gInitial velocity, vi = 1 m/sspring constant, k = 120 N/mCompression, x = 4 cm

We need to calculate the speed and direction of each cart

Step 2. Using Law of conservation of momentum and law of conservation of energy

Firstly,

Using Law of conservation of energy

F_sp = F_net

$$\begin{gathered} kx=ma \\ kx=m\left(v_x\frac{d{v}_x}{dx}\right) \\ kxdx=m{v}_{x}d{v}_x \\ k\int xdx=m\int v_{x}d{v}_x \\ \frac{k{x}^2}{2}=\frac{m{v_x}^2}{2} \\ k{x}^2=m{v_x}^2 \\ Puttingvalues, \\ (120N/m){(0.04m)}^2=(0.3kg){v_x}^2 \\ {v}_x=0.8m/s \end{gathered}$$

Using Law of conservation of Momentum,

$$\begin{gathered} (m_1+m_2)v_i=m_1{v}_{f1}+m_2{v}_{f2} \\ Where{m}_{1}and{m}_{2}aremassesofobject1and2. \\ {v}_{i}istheinitialvelocity. \\ {v}_{f1}and{v}_{f2}arefinalvelocityofobject1and2. \\ Weknow \\ {v}_{f2}=v_i+v_x \\ {v}_{f2}=\left(1m/s+0.8m/s\right) \\ {v}_{f2}=1.8m/s \\ Puttingvaluesinequation, \\ (0.1kg+0.3kg)(1m/s)=(0.1kg)\left(v_{f1}\right)+(0.3kg)(1.8m/s) \\ \left(\frac{0.4-5.4}{0.1}\right)m/s=v_{f1} \\ {v}_{f1}=-1.4m/s \end{gathered}$$

(Negative sign indicates that direction is towards Left.)

Speed of first cart is 1.4 m/s towards Left and speed of second cart is 1.8 m/s towards right.