Question

Two cars collide at an icy intersection and stick together afterward. The 1st car has a mass of 1400kg and was approaching at 7.00m/s due south. The 2nd car has a mass of 900kg and was approaching at 23.0 m/s due west. (a) Calculate the final velocity (magnitude in m/s and direction in degrees counterdockwise from the west) of the cars(B) how much kinetic energy in J is lost in the collision?

Answer 1

(a) magnitude: 9.96 m/s

Direction: 64.67°

(b) 158268.16 Joules

Step-by-step explanation:

Part (a)

We need to apply the conservation of momentum in each direction, so

`\begin{gathered} p_(iy)=p_(fy) \\ m_1v_(iy)=(m_1+m_2)v_(fy) \\ (1400\text{ kg\rparen\lparen7 m/s\rparen= \lparen1400 kg+900 kg\rparen v}_(fy) \end{gathered}`

Solving for vfy, we get:

`\begin{gathered} 9800\text{ kg m/s = 2300 kg v}_(fy) \\ \\ v_(fy)=\frac{9800\text{ kg m/s}}{2300\text{ kg}} \\ \\ v_(fy)=4.26\text{ m/s due to south} \end{gathered}`

For the horizontal direction, we get:

`\begin{gathered} p_(ix)=p_(fx) \\ m_2v_(ix)=(m_1+m_2)v_(fx) \\ (900\text{ kg\rparen\lparen23 m/s\rparen = \lparen1400 kg + 900 kg\rparen}v_(fx) \\ 20700\text{ kg m/s = \lparen2300 kg\rparen}v_(fy) \\ \\ v_(fx)=\frac{20700\text{ kg m/s}}{2300\text{ kg}} \\ \\ v_(fx)=9\text{ m/s} \end{gathered}`

Now, we can calculate the magnitude and direction of the final velocity as follows

`\begin{gathered} v_f=\sqrt{v_(fx)^2+v_(fy)^2} \\ \\ v_f=√(4.26^2+9^2) \\ v_f=9.96\text{ m/s} \\ \\ \\ \theta=\tan^(-1)((v_(fy))/(v_(fx))) \\ \\ \theta=\tan^(-1)((9)/(4.26)) \\ \\ \theta=64.67° \end{gathered}`

Part (b)

Then, the loss in kinetic energy can be calculated as:

`\begin{gathered} \text{ Loss in KE = }KE_i-KE_f \\ \text{ Loss in KE = \lparen}(1)/(2)m_1v_1^2+(1)/(2)m_2v_2^2)-((1)/(2)(m_1+m_2)v_f^2) \\ \\ \text{ Loss in KE = \lparen}(1)/(2)(1400)(7)^2+(1)/(2)(900)(23)^2)-((1)/(2)(1400+900)(9.96)^2) \\ \\ \text{ Loss in KE = 272350 J - 114081.84 J} \\ \text{ Loss in KE = 158,268.16 J} \end{gathered}`

Therefore, the loss in kinetic energy was 158,268.16 Joules.