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Leokris [45]
3 years ago
11

A 70.0-kg astronaut pushes to the left on a spacecraft with a force F in "gravity-free" space. The spacecraft has a total mass o

f 1.0 x 10^4 kg. During the push, the astronaut accelerates to the right with an acceleration of 0.36 m/s^2. Determine the magnitude of the acceleration of the spacecraft.
Physics
1 answer:
goldenfox [79]3 years ago
6 0

Answer:

a_2=-0.00252\ m/s^2

Explanation:

Given that,

Mass of the astronaut, m = 70 kg

Mass of the spacecraft, m'=10^4\ kg

Acceleration of the astronaut, a_1=0.36\ m/s^2

To find,

The magnitude of the acceleration of the spacecraft.

Solution,

The force acting on the astronaut is given by :

F=m\times a_1

F=70\ kg\times 0.36\ m/s^2=25.2\ N

During the push, the astronaut accelerates as a result a reaction force will also act on the spacecraft. Let a_2 is the acceleration of the spacecraft. It can be calculated as :

a_2=\dfrac{-F}{m} (reaction force)

a_2=\dfrac{-25.2}{10^4}

a_2=-0.00252\ m/s^2

Therefore, the magnitude of the acceleration of the spacecraft is 0.00252\ m/s^2. Hence, this is the required solution.

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4 years ago
A 2.0-kg object moving 5.0 m/s collides with and sticks to an 8.0-kg object initially at rest. Determine the kinetic energy lost
densk [106]

Answer:

20J

Explanation:

In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;

m₁u₁ + m₂u₂ = (m₁ + m₂)v          ---------------(i)

Where;

m₁ and m₂ are the masses of first and second objects respectively

u₁ and u₂ are the initial velocities of the first and second objects respectively

v  is the final velocity of the two objects after collision;

From the question;

m₁ = 2.0kg

m₂ = 8.0kg

u₁ = 5.0m/s

u₂ = 0        (since the object is initially at rest)

<em>Substitute these values into equation (i) as follows;</em>

(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v

(10.0) + (0) = (10.0)v

10.0 = 10.0v

v = 1m/s

The two bodies stick together and move off with a velocity of 1m/s after collision.

The kinetic energy(KE₁) of the objects before collision is given by

KE₁ = \frac{1}{2}m₁u₁² +  \frac{1}{2}m₂u₂²       ---------------(ii)

Substitute the appropriate values into equation (ii)

KE₁ = (\frac{1}{2} x 2.0 x 5.0²) +  (\frac{1}{2} x 8.0 x 0²)

KE₁ = 25.0J

Also, the kinetic energy(KE₂) of the objects after collision is given by

KE₂ = \frac{1}{2}(m₁ + m₂)v²      ---------------(iii)

Substitute the appropriate values into equation (iii)

KE₂ = \frac{1}{2} ( 2.0 + 8.0) x 1²

KE₂ = 5J

The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision

K = KE₂ - KE₁

K = 5 - 25

K = -20J

The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J

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