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mihalych1998 [28]

3 years ago

12

Playing catch with a friend in a moving train. When you toss the ball in the direction the train is moving, how does the speed o

f the ball appear to an observer standing at rest outside the train?
I understand basic relative motion, just the ball moving on another moving object confuses me!

2 answers:

likoan [24]3 years ago

5 0

Well it would simply appear faster because the speed of the ball and the speed of the train.

telo118 [61]3 years ago

5 0

Answer:

The speed of the ball appear faster when we observe it from ground frame at rest.

Explanation:

As we know that by relate motion concept

$v_{AB} = v_A - v_B$

here when ball is thrown in the direction of train motion with velocity "v" relative to the train then we will assume here that let say the train at that moment is running at speed $v_T$

now by above concept of relative velocity we will have

$v = v_{ball} - v_T$

now for the speed of ball

$v_{ball} = v + v_T$

so from this equation we can see that the actual speed of ball in ground frame will be more than the speed observed in train frame

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inysia [295]

Answer:

$v = 1,582 \ \frac{m}{s}$

Explanation:

We know that for circular motion the centripetal acceleration $a_c$ is:

$a_c = \frac{v^2}{r}$

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The centripetal acceleration for the astronaut must be the gravitational acceleration due to the gravity, as there are no other force. So

$a_c = 1.27 \frac{m}{s^2}$ .

The radius of the orbit must be the radius of the Moon, plus the 270 km above the surface

$r = 1.7 * 10^6 \ m + 270 \ km$

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$r = 1.97 * 10^6 \ m$

We can obtain the speed as:

$v^2 = a_c r$

$v = \sqrt{a_c r}$

$v = \sqrt{1.27 \frac{m}{s^2} * 1.97 * 10^6 \ m}$

$v = \sqrt{ 2.509 \ 10^6 \ \frac{m^2}{s^2}}$

$v = 1.582 \ 10^3 \ \frac{m}{s}$

$v = 1,582 \ \frac{m}{s}$

And this is the orbital speed.

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3 years ago

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Radda [10]

Answer:

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AleksandrR [38]

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Answer:

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