Question

An astronaut of mass 100 kg, holding a ball of mass 10 kg, is in a space station, at rest
relative to it. He throws the ball with speed 2 m/s relative to himself. What are the speeds
of the astronaut and the ball relative to the space station? Ignore gravity

Answer 1

let velocity of astronaut = $v_a$

velocity of ball = $v_b$

velocity of space station = $v_s$

mass of astronaut = $m_a$ = 100kg

mass of ball = $m_b$ = 10kg

velocity of ball relative to astronaut = $v_{ba}=v_b-va$ = 2 m/s

⇒$v_b=2+v_a$

⇒$v_{bs}=2+v_{as}$.......(1)

using law of conservation of linear momentum,

$m_au_{as}+m_bu_{bs}=m_av_{as}+m_bv_{bs}$

astronaut and ball are in space station and both are in rest relative to space station.

so, $u_{as}=0,u_{bs}=0$

⇒100kg × 0 + 10kg × 0 = 100kg × $v_{as}$ + 10kg × $v_{bs}$

from equation (1),

⇒0 = 100kg × $v_{as}$ + 10kg × $(2+v_{as})$

⇒0 = 100$v_{as}$ + 20 + 10$v_{as}$

⇒$v_{as}$ =-20/110 = -2/11 m/s

so, $v_{bs}$ = 2 - 2/11 = 19/11 m/s

hence, speed of astronaut relative to space station is 2/11 m/s and speed of ball relative to space station is 19/11 m/s just opposite direction of astronaut.

Answer 2

$\huge\bold\purple{Answer:-}$

let velocity of astronaut = v_a

velocity of ball = v_b

velocity of space station = v_s

mass of astronaut = m_a = 100kg

mass of ball = m_b = 10kg

velocity of ball relative to astronaut = v_{ba}=v_b-va = 2 m/s

⇒v_b=2+v_a

⇒v_{bs}=2+v_{as}.......(1)

using law of conservation of linear momentum,

m_au_{as}+m_bu_{bs}=m_av_{as}+m_bv_{bs}

astronaut and ball are in space station and both are in rest relative to space station.

so, u_{as}=0,u_{bs}=0

⇒100kg × 0 + 10kg × 0 = 100kg × v_{as} + 10kg × v_{bs}

from equation (1),

⇒0 = 100kg × v_{as} + 10kg × (2+v_{as})

⇒0 = 100v_{as} + 20 + 10v_{as}

⇒v_{as} =-20/110 = -2/11 m/s

so, v_{bs} = 2 - 2/11 = 19/11 m/s

hence, speed of astronaut relative to space station is 2/11 m/s and speed of ball relative to space station is 19/11 m/s just opposite direction of astronaut.