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miss Akunina [59]
2 years ago
13

Carts A and B are identical and are moving toward each other on a track. The speed of cart A is v, while the speed of cart B is

2v. In the Earth reference frame, the system of the two carts has kinetic energy K.
Physics
1 answer:
borishaifa [10]2 years ago
6 0

Answer: k= \frac{5mv^{2} }{2}

Explanation:

Recall that the formula for kinetic energy is given below as

k = \frac{mv^{2} }{2}

where k=kinetic energy (joules), m= mass of object (kg), v= velocity of object m/s)

For cart A

m_{a} = mass of cart A

v_{a} = v = velocity of cart A

K.E_{a} = kinetic energy of cart A

hence, K.E_{a} = \frac{m_{a}v^{2}  }{2}

For cart B

m_{b} = mass of cart B

v_{b} = 2v = velocity of cart B

K.E_{b} = kinetic energy of cart B

hence, K.E_{b} = \frac{m_{b}(2v^{2}) }{2} = 2m_{b} v^{2}

from the question, both cart are identical which implies they have the same mass i.e m_{a} = m_{b} = m which implies that

K.E_{a}= \frac{mv^{2} }{2} and K.E_{b}  =2mv^{2}

The total kinetic energy K is the sum of cart A and cart B kinetic energy

K=K.E_{a} + K.E_{b}

K=\frac{mv^{2} }{2} + 2mv^{2}

hence

K=\frac{5mv^{2} }{2}

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Answer:

<h2>9.375Nm</h2>

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2 years ago
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A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice
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If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of \sqrt{3} v_{e}

<u>Explanation:</u>

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            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

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From given conditions,

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Similarly, kinetic energy at final would be,

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Now, adding potential and kinetic energies of initial and final and equating as below, find the final velocity as

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'm' and \frac{1}{2} as common on both sides, so gets cancelled, we get as

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We know, v_{e}=\sqrt{\frac{2 G M}{R}}, it can be wriiten as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we get

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking squares out, we get,

                v_{f}=\sqrt{3} v_{e}

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