An astronaut with a mass of #110 kg# is floating in space. If the astronaut throws an object with a mass of #9 kg# at a speed of #5/9 m/s#, how much will his speed change by?

Question

An astronaut with a mass of #110 kg# is floating in space.  If the astronaut throws an object with a mass of #9 kg# at a speed of #5/9 m/s#, how much will his speed change by?

Julia Aguilar · Accepted Answer

#1/22# m/s

Quantities: #m_a = 110# kg #m_o = 9# kg #v_(o_f) = 5/9# m/s #v_(a_f) = ?# m/s #v_(a_i) = v_(o_i) \equiv v_i# m/s #p_(a_i) + p_(o_i) = p_(a_f) + p_(o_f)# #m_av_(a_i) +m_ov_(o_i) = m_av_(a_f) + m_ov_(o_f)# #(m_a + m_o)v_i = m_av_(a_f) + m_ov_(o_f)# #(110 + 9)v_i = (110)v_(a_f) + (9)(5/9)# #119v_i = 110v_(a_f) + 5# Since we are only interested in the change (or difference) between the two speeds, the actual initial and final speeds of the astronaut are irrelevant. For the sake of argument, we can assume that the astronaut's initial speed is zero. After all, it says "floating in space," so perhaps that really means zero. #-5 = 110v_(a_f)# #v_(a_f) = -1/22# This means that the astronauts speed will change by #1/22# m/s and the astronaut will move in the opposite direction of the object (hence the minus sign).

Elena Albarran · Answer

undefined To find the change in the astronaut's speed, we can use the principle of conservation of momentum.

Initial momentum of the system (astronaut + object) = Final momentum of the system

$ m_{astronaut} \times v_{initial} + m_{object} \times v_{object} = (m_{astronaut} + m_{object}) \times v_{final} $

Substituting the given values:

$ 110 \times 0 + 9 \times \frac{5}{9} = (110 + 9) \times v_{final} $

$ \frac{45}{9} = 119 \times v_{final} $

$ v_{final} = \frac{45}{119} $

Therefore, the astronaut's speed will change by $ \frac{45}{119} $ m/s.