Question

3.0kg astronaut throws a 5.0kg
hammer in a direction away from the
shuttle with a speed of 18.Om/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 hamme


​

Answer 1

Given :-

• Mass of Astronaut, $\sf{m_A}$ = 63 kg
• Mass of Hammer, $\sf{m_H}$ = 5 kg
• Initial velocity of Astronaut, $\sf{u_A}$ = 0
• Initial velocity of Hammer, $\sf{u_H}$ = 0
• Final velocity of Hammer, $\sf{v_H}$ = 18 ms⁻¹

To find :-

• Final velocity of Astronaut, $\sf{v_A}$ = ?

⚛ 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,

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

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

• [ 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

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

$\longrightarrow\sf{(63)\;(0)+(5)\;(0)=(63)\;v_A+(5)\;(18)}$

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

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

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

Therefore,

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