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Romashka-Z-Leto [24]
4 years ago
10

Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at

8.00m/s due south. The second car has a mass of 850 kg and is approaching at 17.0m/s due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision
Physics
1 answer:
Andrews [41]4 years ago
3 0

Answer:

a. v=11.73m/s

b. K.E_{lost}=149,202m/s

Explanation:

Since the collision is an inelastic collision, kinetic energy will be reduced and the momentum will be conserved. To calculate the common velocity after collision we use the formula below

m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{1})v\\m_{1}=1200kg, m_{2}=850kg\\v_{1}=8m/s,v_{2}=17m/s\\

If we substitute value into the equation we arrive at

(1200*8)+(850*17)=(1200+850)v\\v=\frac{9600+14450}{2050} \\v=11.73m/s

b. the change in kinetic energy is expressed as

K.E=K.E_{final}- K.E_{initial}\\

K.E_{initial}=1/2m_{1}v_{1}^{2}+1/2m_{2}v_{2}^{2}\\K.E_{initial}=1/2[(1200*8^{2} )+(850*17^{2} )]\\K.E_{initial}=1/2(76800+245650)\\K.E_{initial}=161225m/s

K.E_{final}=1/2(m_{1}+m_{2})v\\K.E_{final}=1/2(1200+850)11.73\\K.E_{final}=12,023m/s

The lost kinetic energy is calculated as

K.E_{lost}=K.E_{final}- K.E_{initial}\\K.E_{lost}=12023-161225\\K.E_{lost}=149,202m/s

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