Question

A 1420-kg car moving east at 17.0 m/s collides with a 1880-kg car moving south at 15.0 m/s, and the two cars connect together.A. What is the magnitude of the velocity of the cars right after the collision? (m/s )B. What is the direction of the cars right after the collision? Enter the angle in degrees where positive indicates north of east and negative indicates south of east. (°)C. How much kinetic energy was converted to another form during the collision? (kJ)

Answer 1

Given,

The mass of the car moving east, m=1420 kg

The mass of the car moving south, M=1880 kg

The velocity of the car moving east, u₁=17.0 m/s

The velocity of the car moving south, u₂=-15.0 m/s

Here we assume that the eastward direction is the positive x-direction and the southward direction is the negative y-direction.

From the law of conservation of momentum, the momentum is conserved in both directions simultaneously and independently.

Considering the conservation of momentum in the x-direction,

$mu_1=(m_{}+M)v_x$

Where v_x is the x-component of the final velocity of the two cars.

On substituting the known values,

$\begin{gathered} 1420\times17.0=(1420+1880)v_x \\ v_x=\frac{1420\times17.0}{(1420+1880)} \\ =7.32\text{ m/s} \end{gathered}$

Considering the conservation of momentum in the y-direction,

$Mu_2=(m+M_{})v_y$

Where v_y is the y-component of the final velocity of the cars.

On substituting the known values,

$\begin{gathered} _{}1880\times-15.0=(1420+1880)v_y \\ v_y=\frac{1880\times-15.0}{(1420+1880)} \\ =-8.55\text{ m/s} \end{gathered}$

A.

The magnitude of the velocity of the cars tight after the collision is given by,

$v=\sqrt[]{v^2_x+v^2_y}$

On substituting the known values,

$\begin{gathered} v=\sqrt[]{7.32^2+(-8.55)^2} \\ =11.26\text{ m/s} \end{gathered}$

Thus the magnitude of the velocity of the cars right after the collision is 11.25 m/s

B.

The direction of the cars right after the collision is given by,

$\theta=\tan ^{-1}(\frac{v_y}{v_x})$

On substituting the known values,

$\begin{gathered} \theta=\tan ^{-1}(\frac{-8.55}{7.32}) \\ =-49.4^{\circ} \end{gathered}$

Thus the direction of the cars right after the collision is -49.4°. That is 49.4° south of the east.

C.

The total kinetic energy of the system is before the collision is

$K_1=\frac{1}{2}mu^2_1+\frac{1}{2}Mu^2_2$

On substituting the known values,

$\begin{gathered} K_1=\frac{1}{2}\times1420\times17.0^2+\frac{1}{2}\times1880\times15.0^2 \\ =205.19\times10^3+211.5\times10^3 \\ =416.69\times10^3\text{ J} \end{gathered}$

The kinetic energy of the system of two cars after the collision is,

$\begin{gathered} K_2=\frac{1}{2}(m+M)v^2 \\ =\frac{1}{2}\times(1420+1880)11.26^2 \\ =209.2\times10^3\text{ J} \end{gathered}$

Thus the kinetic energy lost during the collision is,

$\Delta K_{}=K_1-K_2$

On substituting the known values,

$\begin{gathered} \Delta K=416.69\times10^3-209.2\times10^3 \\ =207.49\times10^3\text{ J} \\ \approx207.5\text{ kJ} \end{gathered}$

Thus the total kinetic energy lost during the collision is 207.5 kJ