Respuesta :
(a) 0.165 m/s
The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):
[tex]p_i = 0[/tex]
The final total momentum is instead:
[tex]p_f = m_a v_a + m_c v_c[/tex]
where
[tex]m_a = 125 kg[/tex] is the mass of the astronaut
[tex]v_a = 2.50 m/s[/tex] is the velocity of the astronaut
[tex]m_c = 1900 kg[/tex] is the mass of the capsule
[tex]v_c[/tex] is the velocity of the capsule
Since the total momentum must be conserved, we have
[tex]p_i = p_f = 0[/tex]
so
[tex]m_a v_a + m_c v_c=0[/tex]
Solving the equation for [tex]v_c[/tex], we find
[tex]v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s[/tex]
(negative direction means opposite to the astronaut)
So, the change in speed of the capsule is 0.165 m/s.
(b) 520.8 N
We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:
[tex]F \Delta t = \Delta p[/tex]
The change in momentum of the astronaut is
[tex]\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s[/tex]
And the duration of the push is
[tex]\Delta t = 0.600 s[/tex]
So re-arranging the equation we find the average force exerted by the capsule on the astronaut:
[tex]F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N[/tex]
And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.
(c) 25.9 J, 390.6 J
The kinetic energy of an object is given by:
[tex]K=\frac{1}{2}mv^2[/tex]
where
m is the mass
v is the speed
For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is
[tex]K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J[/tex]
For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is
[tex]K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J[/tex]
(a)The change in the speed of the space capsule will be 0.165 m/sec.
(b)The average force exerted by each on the other will be 520.8 N.
(c)The kinetic energy of each after the push for astronaut and capsule will be 25.9 J and 390.6 J respectively.
What is speed?
Speed is defined as the rate of change of the distance or the height attained. it is a time-based quantity. it is denoted by u for the initial speed while v for the final speed. its si unit is m/sec.
(a)The change in the speed of the space capsule will be 0.165 m/sec.
The initial momentum of the astronaut-capsule system is zero due to rest.
[tex]\rm P_f=\rm m_av_a+m_cv_c \\\\ \rm P_I=0 \\\\ \rm m_av_a+m_cv_c =0 \\\\ \rm v_c=\frac{-m_av_a}{m_c} \\\\ \rm v_c=\frac{-125\times 2.50 }{1900} \\\\ \rm v_c=-0.165[/tex]
Hence the changes in the speed of the space capsule will be 0.165 m/sec.
(b)The average force exerted by each on the other will be 520.8 N.
According to the impulse-momentum theorem
[tex]\rm F\triangle t = \triangle P \\\\ \triangle P= m\triangle v \\\\ \triangle P= 125\times 250 =312\ Kgm/sec.[/tex]
t is the time interval = 0.600 s
[tex]\rm F=\frac{\triangle p}{\triangle t} \\\\\rm F=\frac{\312 }{0.600} \\\\\rm F= 520.8 \N[/tex]
Hence the average force exerted by each on the other will be 520.8 N.
(c)The kinetic energy of each after the push for astronaut and capsule will be 25.9 J and 390.6 J respectively.
The kinetic enrgy is given by the formula;
The kinetic energy of the astronaut is;
[tex]\rm KE=\frac{1}{2}mv^2 \\\\ \rm KE=\frac{1}{2}125 \times (250)^2 \\\\ \rm KE=390 J[/tex]
The kinetic energy of the capsule is;
[tex]\rm KE=\frac{1}{2}mv^2 \\\\ \rm KE=\frac{1}{2}1900 \times (0.165)^2 \\\\ \rm KE=25.9 J[/tex]
Hence the kinetic energy of each after the push for astronaut and capsule will be 25.9 J and 390.6 J respectively.
To learn more about the speed refer to the link;
https://brainly.com/question/7359669
