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erastovalidia [21]
3 years ago
11

A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the s

ame road at 14.0 m/s . The two vehicles remain locked together after the collision.
a. What is the velocity (magnitude and direction) of the two vehicles just after the collision?
b. At what speed should the truck have been moving so that both it and the car are stopped in the collision?
c. Find the change in kinetic energy of the system of two vehicles for the situations of part (a) and part (b). For which situation is the change in kinetic energy greater in magnitude?

Physics
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

(A) V = 9.89m/s

(B) U = -2.50m/s

(C) ΔK.E = –377047J

(D) ΔK.E = –257750J

Explanation:

The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.

This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.

See attachment below for full solution.

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3. A large crane lifts a 25,000 kg mass in the air. The amount of work that must be done by the
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\mathfrak{\huge{\orange{\underline{\underline{AnSwEr:-}}}}}

Actually Welcome to the concept of Efficiency.

Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%

The efficiency is => 22% => 22/100.

so we get as,

E = W(output) /W(input)

hence, W(output) = E x W(input)

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Here we have mass that moves at ceratin speed. This means that we have momentum. The law that must be observed is law of conservation of momentum. It states that momentum before certain event is equal to a momentum after that event. Here we have three masses so we can write this as:
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Before the firecracker blows a coconut does not move, so left side is equal to 0:
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We know that m1=m2=m and m3=2m. Also we are asked to find v3f so we can rewrite formula:
v_{3f} = -  \frac{m_{1}  v_{1f}  + m_{2} v_{2f} }{ m_{3} }

We must take in consideration that two parts with same mass do not move in same direction. The center of mass of these two parts moves between them at angle of 45° with respect to both south and west. The speed of a center of mass is:
v_{f} = \sqrt{ v_{1f}^{2}+ v_{2f}^{2} } \\ \\ v_{f} = 33.9m/s
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v_{3f} = - \frac{m*33.9+m*33.9 }{ 2m } \\  \\ v_{3f} = - \frac{2m*33.9 }{ 2m }  \\  \\ v_{3f} = - 33.9m/s

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