Find the percentage of the total work lost to friction if 28.7 J of work is put into pushing a block up a ramp resulting in 14.2
1 answer:
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The collision of car 2 is more violent (because more impulse is exerted)
Explanation:
The collision which is more violent is the one in which more impulse is exchanged.
The impulse exerted by each car on the wall is equal to the change in momentum of the car:
where
m is the mass of the car
is the change in velocity
For the car 1,
m = 1000 kg
(the sign is negative because the velocity of the car has changed from 100 km/h to 0 km/h)
So the magnitude of the impulse of car 1 is
For the car 2,
m = 1200 kg
So the magnitude of the impulse of car 2 is
So, car 2 exerts a larger impulse, therefore its impact is more violent.
Learn more about impulse and change in momentum:
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