A motionless 120 kg astronaut is holding a 14 kg tool while on a spacewalk. To get moving, the astronaut throws the tool forward at a speed of +6.4 m/s. How fast does the astronaut move backward?
To solve this problem, we can apply the conservation of momentum principle.
According to the conservation of momentum, the total momentum before an event is equal to the total momentum after the event (assuming no external forces act on the system).
The momentum of an object is given by the equation:
Momentum = mass * velocity
Initially, both the astronaut and the tool are motionless. Therefore, the initial momentum of the system is zero.
After the astronaut throws the tool forward, there are two objects with non-zero velocities. Let the velocity of the tool be v_tool and the velocity of the astronaut be v_astronaut (both in the backward direction).
The momentum after the event is given by:
Momentum_after = (mass_astronaut * v_astronaut) + (mass_tool * v_tool)
We know the mass of the astronaut (120 kg), the mass of the tool (14 kg), and the velocity of the tool (+6.4 m/s).
Plugging in the values into the equation, we get:
0 = (120 kg * v_astronaut) + (14 kg * +6.4 m/s)
Simplifying the equation, we get:
0 = 120 kg * v_astronaut + 89.6 kg*m/s
Dividing both sides of the equation by 120 kg, we get:
0 = v_astronaut + 0.7467 m/s
Rearranging the equation to solve for v_astronaut, we get:
v_astronaut = -0.7467 m/s
Therefore, the astronaut moves backward at a speed of approximately 0.75 m/s.