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koban [17]
3 years ago
15

Two astronauts in space with a baseball decide to play catch to pass the time. In the language of conservation of momentum, desc

ribe what happens to each astronaut as they start to toss the ball back and forth
Physics
2 answers:
IceJOKER [234]3 years ago
7 0

Answer:  Suppose that the first astronaut is still in place, then the full momentum of that astronaut is zero.

Now, when the astronaut throws the ball, now the ball has momentum, so the astronaut moves in the opposite direction to conserve the momentum (the movement of the arm also creates a response in the body of the astronaut)

(all of this can be explained also by the third Newton's law, for example, the astronaut that accelerates the baseball also experiences a force that the baseball does in him)

Usually, in the earth, the force of gravity keeps the players in place, but in the space, this is not the case, so the tiny force that the ball does in the astronaut is enough to accelerate the astronaut.

It is the same for the other one, the ball comes with a little bit of momentum, so when he catches the ball, the momentum must be conserved, so the astronaut will move in the same direction that the ball was moving.

Anna35 [415]3 years ago
4 0
As the first astronaut throws the ball, lets assume it goes with v velocity and the mass of the ball be m
the momentum comes out be mv, thus to conserve that momentum the astronaut will move opposite to the direction of the ball's motion with the velocity mv/M (where M is the mass of the astronaut).
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F= 17,075\ N

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3 0
3 years ago
In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e
Zanzabum

Answer:

The  velocity is  v_t  =  0.02175 \  m/s

Explanation:

From the question we are told that

   The  mass of the bullet is  m_b  =  0.024 \  kg

    The initial speed of the bullet is  u_b  =  1200 \  m/s

   The mass of the target is  m_t  =  320 \  kg

    The  initial velocity of target is  u_t  =  0  \ m/s

    The  final velocity of the bullet is  is  v_b  =  910 \  m/s

   

Generally according to the law of momentum conservation we have that

      m_b *  u_b  +  m_t *  u_t  =  m_b *  v_b  +  m_t  *  v_t

=>   0.024  *  1200  +  320 *  0  =  0.024 *  910   +  320  *  v_t

=>    v_t  =  0.02175 \  m/s

3 0
3 years ago
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