Question

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1750 kg and was approaching at 6.00 m/s due south. The second car has a mass of 700 kg and was approaching at 19.0 m/s due west.(a) Calculate the final velocity of the cars. (Note that since both cars have an initial velocity, you cannot use the equations for conservation of momentum along the x-axis and y-axis; instead, you must look for other simplifying aspects..)Magnitude m/sDirection ° (counterclockwise from west is positive)(b) How much kinetic energy is lost in the collision? (This energy goes into deformation of the cars.) J

Answer 1

Given:

• Mass of first car 1, m1 = 1750 kg

,

• Mass of second car, m2 = 700 kg

,

• Initial velocity of car 1, u1 = 6.00 m/s due south

,

• Initial velocity of car 2, u2 = 19.0 m/s due west.

Let's solve for the following:

• (a). Calculate the final velocity of the cars.

Apply the conservation of momentum.

In the West direction(x-direction), we have:

$\begin{gathered} m_1u_1+m_2u_2=(m_1+m_2)v_x \\ \\ 1750(0)+700(19.0)=(1750+700)v_x \\ \\ 13300=2450v_x \\ \\ v_x=\frac{13300}{2450} \\ \\ v_x=5.43\text{ m/s} \end{gathered}$

In the South direction (y-direction), we have:

$\begin{gathered} m_1u_1+m_2u_2=(m_1+m_2)v_y \\ \\ 1750(6.00)+700(0)=(1750+700)v_y \\ \\ 10500=2450v_y \\ \\ v_y=\frac{10500}{2450} \\ \\ v_y=4.29\text{ m/s} \end{gathered}$

To find the magnitude, we have:

$\begin{gathered} v=\sqrt{v_x^2+v_y^2} \\ \\ v=\sqrt{5.43^2+4.29^2} \\ \\ v=\sqrt{29.4849+18.4041} \\ \\ v=\sqrt{47.889} \\ \\ v=6.92\text{ m/s} \end{gathered}$

To find the direction, apply the formula:

$\begin{gathered} \theta=tan^{-1}(\frac{v_y}{v_x}) \\ \\ \theta=tan^{-1}(\frac{4.29}{5.43}) \\ \\ \theta=tan^{-1}(0.79) \\ \\ \theta=38.31^o\text{ south of west} \end{gathered}$

• (b). How much kinetic energy is lost in the collision?

To find the loss of kinetic energy, apply the formula:

$KE_{loss}=KE_{initial}-KE_{final}$

Using the Kinetic energy formula, we have:

$\begin{gathered} KE_{loss}=(\frac{1}{2}m_1u_1^2+\frac{1}{2}m_2u_2^2)-\frac{1}{2}(m_1+m_2)v^2 \\ \\ KE_{loss}=(\frac{1}{2}*1750*6.00^2+\frac{1}{2}*700*19^2)-(\frac{1}{2}(1750+700)6.92^2) \end{gathered}$

Solving further:

$\begin{gathered} KE_{loss}=(31500+126350)-58660.84 \\ \\ KE_{loss}=157850-58660.84 \\ \\ KE_{loss}=99189.16\text{ J} \end{gathered}$

Therefore, the amount of kinetic energy lost is 99189.16 J.

• ANSWER:

(a). Magnitude = 6.92 m/s

Direction = 38.31° South of West

(b). 99189.16 J.