Question

A cart with mass 340 gmoving on a frictionless linear air track at an initial speed of 1.2 m/sundergoes an elastic collision with an initially stationary cart of unknown mass. After the collision, the first cart continues in its original direction at 0.66 m/s. (a) What is the mass of the second cart? (b) What is its speed after impact? (c) What is the speed of the two-cart center of mass?

Short answer

a) Mass of the second cart is 0.0987 kg

b) Speed after impact is 1.9 m/s

c) The speed of the two – cart center of mass is 0.93 m/s

Step-by-step solution

Listing the given quantities

Mass of the first cart is $m_1=340g$

$v_{1i}=1.2m/s$

$v_{1f}=0.66m/s$

Understanding the concept of conservation of the linear momentum

We have a moving cart colliding with the stationary cart. Since the collision is elastic, the total kinetic energy remains unchanged.

Let m₁ be the mass of the cartthat is originallymoving${\mathit{v}}_{\mathbf{1}\mathit{i}}$be its velocity before the collision and${\mathit{v}}_{\mathbf{1}\mathbf{f}}$be its velocity after the collision. Let m₂ be the mass of the cart that is originally at rest${\mathit{v}}_{2f}$be its velocity after the collision. Conservation of linear momentum gives${\mathit{m}}_{\mathbf{1}}{\mathit{v}}_{\mathbf{u}}\mathbf{=}{\mathit{m}}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}\mathbf{f}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathit{v}}_{\mathbf{2}\mathbf{f}}$

Similarly total energy is conserved gives $\frac{\mathbf{1}}{\mathbf{2}}{\mathit{m}}_{\mathbf{1}}{\left(v_{1i}\right)}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{m}}_{\mathbf{1}}{\mathbf{\left(}{\mathbf{v}}_{\mathbf{1}\mathbf{f}}\mathbf{\right)}}^{\mathbf{2}}\mathbf{+}{\mathit{m}}_{\mathbf{2}}{\mathbf{\left(}{\mathbf{v}}_{\mathbf{2}\mathbf{f}}\mathbf{\right)}}^{\mathbf{2}}$

Formula used

Solving for$v_{1f}$ and$v_{2f}$ we obtain

$v_{1f}=\frac{m_1-m_2}{m_1-m_2}{v}_{1i},v_{2f}=\frac{2m_1}{m_1+m_2}{v}_{1i}$

The speed of the centre of mass is$v_{com}=\frac{m_1{v}_{1i}-m_2{v}_{2i}}{m_1=m_2}$

 Calculation of the mass of the second cart

(a)

$$\begin{gathered} m_2=\frac{v_{1f}-v_{1f}}{v_{1f}+v_{1f}}{m}_1, \\ =\left(\frac{1.2m/s-0.66m/s}{1.2m/s+0.66m/s}\right)\left(0.34kg\right) \\ =0.0987kg \end{gathered}$$

Mass of the second cart is 0.0987 kg

 Calculation of the velocity of the second cart

(b)

$$\begin{gathered} m_{2f}=\frac{2m_1}{m_1+m_2}{v}_{1i}, \\ =\frac{2(0.34kg)}{0.34kg+0.099kg}\left(1.2m/s\right) \\ =1.9m/s \end{gathered}$$

Speed after impact is 1.9 m/s

Calculation of the speed of the two – cart center of mass

(c)

$$\begin{gathered} v_{com}=\frac{m_1{v}_{1i}-m_2{v}_{2i}}{m_1+m_2} \\ =\frac{(0.34kg)(1.2m/s)+0}{0.34kg+0.099} \\ =0.93m/s \end{gathered}$$

The speed of the two – cart center of mass is 0.93 m/s.