Question

A 126- kg astronaut (including space suit) acquires a speed of 2.70 m/s by pushing off with her legs from a 1800-kg space capsule. Use the reference frame in which the capsule is at rest before the push.
a)What is the velocity of the space capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units. Enter positive value if the direction of the velocity is in the direction of the velocity of the astronaut and negative value if the direction of the velocity is in the direction opposite to the velocity of the astronaut.
b)If the push lasts 0.600 s , what is the magnitude of the average force exerted by each on the other?
Express your answer to three significant figures and include the appropriate units.
c)What is the kinetic energy of the astronaut after the push in the reference frame?
Express your answer to three significant figures and include the appropriate units.
d)What is the kinetic energy of the capsule after the push in the reference frame?
Express your answer to two significant figures and include the appropriate units.

Answer 1

The change in the speed of the space capsule will be -0.189 m/s.

The average force exerted by each on the other will be 567 N.

The kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Given:

Mass of the astronaut, $m_a$ = 126 kg

Speed he acquires, $v_{a}$  = 2.70 m/s

Mass of the space capsule, $m_{c}$ = 1800kg

The initial momentum of the astronaut-capsule system is zero due to rest.

$P_f = m_av_a + m_cv_c$

$P_I$ = 0

$m_av_a + m_cv_c = 0$

$v_c =\frac{- m_a v_a}{m_c}}$

   $= \frac{126* 2.70}{1800}$

   $= - 0.189$ m/s

Therefore,

According, to the impulse-momentum theorem;

FΔt = ΔP

ΔP = m Δv

ΔP = 126×2.70

    = 340.2 kgm/sec

t is time interval = 0.600s

F = ΔP/Δt

F = 340.2/0.600

  = 567 N

Therefore, the average force exerted by each on the other will be 567 N.

The Kinetic Energy of the astronaut;

K.E = $\frac{1}{2} m v^2$

     $= \frac{1}{2}$ × 126 × $(2.70) ^2$

     = 459.27 J

The Kinetic Energy of the capsule;

K.E = $\frac{1}{2} m v^2$

     = $\frac{1}{2}$×1800×$(0.189) ^2$

     = 32.14 J

Therefore, the kinetic energy of each after the push for the astronaut and the capsule are 459.27 J and 32.14 J.

Learn more about kinetic energy here:

brainly.com/question/26520543

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