Question

A 63.0kg astronaut throws a 5.0kg hammer in a direction away from the shuttle with a speed of 18.0m/s, pushing the astronaut back to the shuttle. Assuming that the astronaut and hammer start from rest, find the final speed of the astronaut after throwing the hammer.​

Answer 1

Answer:

The law of conservation of momentum states when a system of interacting objects is ... A 63.0kg astronaut throws a 5.0kg hammer in a direction away from the shuttle with a speed of 18.0m/s, pushing the astronaut back to the shuttle. ... hammer start from rest, find the final speed of the astronaut after throwing the hammer.

Explanation:

Answer 2

Question

A 63.0kg astronaut throws a 5.0kg hammer in a direction away from the shuttle with a speed of 18.0m/s, pushing the astronaut back to the shuttle. Assuming that the astronaut and hammer start from rest, find the final speed of the astronaut after throwing the hammer.

Given :

Mass of Astronaut, $\sf{m_A}$

Mass of Hammer, $\sf{m_H}$

Initial velocity of Astronaut, $\sf{u_A}$

Initial velocity of Hammer, $\sf{u_H}$

Final velocity of Hammer, $\sf{v_H}$

To find :

Final velocity of Astronaut, $\sf{v_A}v$

Knowledge required :

Law of conservation of momentum

The law of conservation of momentum states when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change. i.e,

$\implies\boxed{\textsf{total initial momentum= total final momentum}}⟹$

total initial momentum= total final momentum

$\implies\boxed{\sf{m_1\;u_1+m_2\;u_2=m_1\;v_1+m_2\;v_2}}⟹$

[ where m₁ and m₂ are mass of two bodies, u₁ and u₂ are initial velocity of two bodies and v₁ and v₂ are final velocities of two bodies ]

Solution :

Using Law of conservation of momentum

$\longrightarrow\sf{m_A\;u_A+m_H\;u_H=m_A\;v_A+m_H\;v_H}⟶m$

$\longrightarrow\sf{(63)\;(0)+(5)\;(0)=(63)\;v_A+(5)\;(18)}⟶(63)(0)+(5)(0)=(63)v </p><p>A$

$\longrightarrow\sf{0=63\;v_A+90}⟶0=63v$

$\longrightarrow\sf{-90=63\;v_A}⟶−90=63v$

$\longrightarrow\underline{\underline{\red{\sf{v_A=-1.4285\;\;ms^{-1}}}}}$

Therefore,

Final velocity of Astronaut will be 1.4285 ms⁻¹ approximately, In opposite direction to the motion of Hammer.