Question
Asked Oct 24, 2019
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A 1,195-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9,600-kg truck moving in the same direction at 20.0 m/s (see figure below). The velocity of the car right after the collision is 18.0 m/s to the east.

a. What is the velocity of the truck right after the collision? (Round your answer to at least three decimal places.)

b.  How much mechanical energy is lost in the collision? Account for this loss in energy.

+25.0 m/s
+20.0 m/s
+18.0 m/s
Before
After
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+25.0 m/s +20.0 m/s +18.0 m/s Before After

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Expert Answer

Step 1

a)

According to momentum conservation law,

n = meg'fcx + MpusViruck
maVic
i trock
(1195kg)(25.0m/s) (9600 kg ) ( 20.0 m/s) = (1195 kg) (18.0m/s) + (9600kg)v; de
2.21x10 kg m/s 0.215x10 kg -m/s+(9600 kg ) ,;eeck
(2.21-0.215)x 10 kg -m/s
trudk
ftroc
9600kg
20.8m/s
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n = meg'fcx + MpusViruck maVic i trock (1195kg)(25.0m/s) (9600 kg ) ( 20.0 m/s) = (1195 kg) (18.0m/s) + (9600kg)v; de 2.21x10 kg m/s 0.215x10 kg -m/s+(9600 kg ) ,;eeck (2.21-0.215)x 10 kg -m/s trudk ftroc 9600kg 20.8m/s

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Step 2

Answer: 20.8 m/s

Step 3

b)

Mechanical energy lost i...

Et(1195kg(25.0m/s)- (18.0m/s)(9600 kg) ( (20.0 m/s)- (20.8m/s))
=0.5(359,695 J-313,344 J
1kJ
=46,351
10 J
46.3 kJ
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Et(1195kg(25.0m/s)- (18.0m/s)(9600 kg) ( (20.0 m/s)- (20.8m/s)) =0.5(359,695 J-313,344 J 1kJ =46,351 10 J 46.3 kJ

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Work,Power and Energy

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