A 1000-kg car is traveling east at 20\:m.s^{-1}20 m . s − 1 and a 1200-kg car is traveling west at 22\:m.s^{-1}\:22 m . s − 1. W
1 answer:
Answer:
490,400 J
Explanation:
Mass of first car, m = 1000 kg
Mass of second car, M = 1200 kg
velocity of first car, u = 20 m/s east
velocity of second car, U = 22 m/s west
The formula for the kinetic energy is
where, m is the body and v be the velocity of the body.
Total kinetic energy is given by
k = 490,400 J
Thus, the total kinetic energy of the system is 490,400 J.
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a) 5.0 m/s
This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:
where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.
When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:
where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find
b) 4.0 m/s
After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:
where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find
c) 2.8 m
We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:
While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:
where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find