Question

A 125-kg astronaut (including space suit) acquires a speed of \({\bf{2}}{\bf{.50}}\;{{\bf{m}} \mathord{\left/{\vphantom {{\bf{m}} {\bf{s}}}} \right.\\} {\bf{s}}}\) by pushing off with her legs from a 1900-kg space capsule.

(a) What is the change in speed of the space capsule?

(b) If the push lasts 0.600 s, what is the average force exerted by each on the other? As the reference frame, use the position of the capsule before the push. (c) What is the kinetic energy of each after the push?

Short answer

(a) The change in the speed of the space capsule is \(0.164\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).

(b) The force exerted by the astronaut on the capsule or by the capsule on the astronaut is\(521\;{\rm{N}}\).

(c)The kinetic energies of the astronaut and the capsule after the push are\(391\;{\rm{J}}\)and\(26\;{\rm{J}}\), respectively.

Step-by-step solution

Define momentum

The momentum of a particle can be obtained by multiplying its velocity with its mass. It varies directly with the particle’s velocity.

Given information

The mass of the astronaut is\({m_1} = 125\;{\rm{kg}}\).

The final velocity of the astronaut is\({v_{1,f}} = 2.50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

The mass of the space capsule is\({m_2} = 1900\;{\rm{kg}}\).

The time is\(t = 0.600\;{\rm{s}}\).

Since the system is at rest before the push, the initial velocity of the astronaut and the space capsule is \({v_{1,{\rm{i}}}} = {v_{2,{\rm{i}}}} = 0\).

Calculate the change in the speed of the space capsule

(a)

Apply the law of conservation of linear momentum to calculate the final speed of the space capsule.

\(\begin{array}{c}{m_1}{v_{1,{\rm{i}}}}+{m_2}{v_{2,{\rm{i}}}}={m_1}{v_{1,f}}+ {m_2}{v_{2,{\rm{f}}}}\\\left({125\;{\rm{kg}}}\right)\left(0\right)+\left( {1900\;{\rm{kg}}} \right)\left(0\right)= \left( {125\;{\rm{kg}}} \right)\left( {2.50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right) + \left( {1900\;{\rm{kg}}} \right){v_{2,{\rm{f}}}}\\{v_{2,{\rm{f}}}}=-\frac{{\left({125\;{\rm{kg}}}\right)\left( {2.50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)}}{{\left({1900\;{\rm{kg}}}\right)}}\\{v_{2,{\rm{f}}}}=-0.164\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\end{array}\)

The change in the speed of the space capsule can be calculated as shown below.

\(\begin{array}{c}\Delta{v_2}={v_{2,{\rm{f}}}}-{v_{2,i}}\\\Delta{v_2}=\left({- 0.164\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}} \right)-0\\\Delta{v_2}=-0.164\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}} \right.\\} {\rm{s}}}\end{array}\)

Here, a negative sign indicates that the speed of the capsule is in the opposite direction to that of the astronaut.

Thus, the change in the speed of the space capsule is\(0.164\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {\rm{s}}}} \right.\\} {\rm{s}}}\).

Calculate the force exerted by the astronaut on the capsule and by the capsule on the astronaut 

(b)

The force exerted by the astronaut on the capsule can be calculated as shown below:

\(\begin{array}{c}{F_1}=\frac{{{m_1}\left({{v_{2,{\rm{f}}}}{v_{1,{\rm{i}}}}}\right)}}{t}\\{F_1}=\frac{{\left({125\;{\rm{kg}}}\right)\left[{\left({2.50\;{{\rm{m}}\mathord{\left/{\vphantom{{\rm{m}}{\rm{s}}}}\right.\\}{\rm{s}}}}\right)\left(0\right)}\right]}}{{\left({0.600\;{\rm{s}}}\right)}}\\{F_1}=521\;{\rm{N}}\end{array}\)

According to Newton’s third law, every action is followed by an opposite reaction. Therefore, the force exerted by the astronaut on the capsule is equal and opposite to the force exerted by the capsule on the astronaut.

Thus, the force exerted by the astronaut on the capsule and by the capsule on the astronaut is\(521\;{\rm{N}}\).

Calculate the kinetic energy of the astronaut and the capsule after the push

(c)The kinetic energy of the astronaut after the push can be calculated as shown below:

\(\begin{array}{l}{K_1} = \frac{1}{2}{m_1}v_{1,{\rm{f}}}^2\\{K_1} = \frac{1}{2}\left( {125\;{\rm{kg}}} \right){\left( {2.50\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)^2}\\{K_1} = 391\;{\rm{J}}\end{array}\)

The kinetic energy of the capsule after the push can be calculated as shown below:

\(\begin{array}{l}{K_2} = \frac{1}{2}{m_2}v_{2,{\rm{f}}}^2\\{K_2} = \frac{1}{2}\left( {1900\;{\rm{kg}}} \right){\left( { - 0.164\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}} \right)^2}\\{K_2} = 26\;{\rm{J}}\end{array}\)

Thus, the kinetic energies of the astronaut and the capsule after the push are \(391\;{\rm{J}}\) and \(26\;{\rm{J}}\), respectively.