Question

problem 4: an astronaut of mass ( m = 120 ) kg is standing stil in space and throws a ball. the mass of the ball is ( m = 200 ) g. the final velocity of the ball after being thrown is ( v_{b}=-10 ) m/s. use conservation of momentum to find the final velocity ( (v_{a}) ) of the astronaut.

Explanation:

Step1: Convert the mass of the ball to kg

Since \(1\space kg = 1000\space g\), then \(m=200\space g=\frac{200}{1000}\space kg = 0.2\space kg\)

Step2: Apply the law of conservation of momentum

The initial momentum \(p_i = 0\) (because the astronaut and the ball are initially at rest, \(v_{i,a}=v_{i,b} = 0\)). According to the law of conservation of momentum \(p_i=p_f\), where \(p_f = Mv_a+mv_b\)
So, \(0 = Mv_a+mv_b\)

Step3: Solve for \(v_a\)

Rearrange the equation \(Mv_a+mv_b = 0\) to get \(v_a=-\frac{mv_b}{M}\)
Substitute \(M = 120\space kg\), \(m = 0.2\space kg\) and \(v_b=- 10\space m/s\) into the formula:
\(v_a=-\frac{0.2\times(-10)}{120}=\frac{2}{120}=\frac{1}{60}\approx0.0167\space m/s\)

Answer:

The final velocity of the astronaut \(v_a\approx0.0167\space m/s\)