Question

Two astronauts floating at rest with respect to their ship in space decide to play catch with a 2-kg asteroid. Joe (whose mass is 100 kg) throws the asteroid at 20 m/s toward Jesse (whose mass is 68 kg). She catches it and throws it back at 18 m/s (with respect to the ship). What is Joe's speed right after his first throw? m/s What is Jesse's speed right after her first catch? m/s What is Jesse's speed right after her first throw? m/s What is Joe's speed right after his first catch? m/s

Answer 1

Mass of Joe = m_Joe = 100 kg

Mass of Asteroid = m_asteroid = 2 kg

Mass of Jesse = m_Jesse = 68 kg

Speed of Joe = v_Joe

Speed of Asteroid = v_asteroid

Speed of Jesse = v_Jesse

Initially all the three bodies are at rest relative to the space ship. So, the total linear momentum is zero.

Hence, throughout the catching and throwing the total momentum of the three body system would be zero, that is,

(m_Joe) (v_Joe) + (m_asteroid) (v_asteroid )+ (m_Jesse) (v_Jesse) = 0 …………………(Equation 1)

Using the above equation for Joe’s 1^st throw, we have,

( 100 kg) (v_Joe) + ( 2 kg ) (20 m/s) + ( 68 kg) (0) = 0 [Since Jesse is still at rest, so v_Jesse = 0]

=> ( 100 kg) (v_Joe) + 40 kg m/s = 0

=> ( 100 kg) (v_Joe) = - 40 kg m/s

=> v_Joe =

=> v_Joe = - 0.4 m/s

Therefore, speed of Joe just after 1^st throw would be 0.4 m/s away from Jesse and speed of asteroid is 20 m/s towards Jesse.

Again, using equation 1 for Jesse’s 1^st catch, we have,

( 100 kg) (- 0.4 m/s) + ( 2 kg ) (v_asteroid )+ ( 68 kg) (v_Jesse) = 0

Now both the asteroid and Jesse would have the same speed, v as they can be considered as a single entity.

Therefore, the above equation reduces to,

- 40 kg m/s + ( 2 kg ) (v)+ ( 68 kg) (v) = 0

=> - 40 kg m/s + (70 kg ) (v) = 0

=> (70 kg ) (v) = 40 kg m/s

=> v =

=> v = 4/7 m/s

=> v = 0.57 m/s

So, the speed of both Jesse and asteroid is 0.57 m/s away from Joe

Again using equation 1 for Jesse’s 1^st throw, we have,

( 100 kg) (- 0.4 m/s) + ( 2 kg ) (- 18 m/s )+ ( 68 kg) (v_Jesse) = 0 [ -18 m/s for asteroid as it is moving towards Joe]

=> -40 kg m/s -36 kg m/s + ( 68 kg) (v_Jesse) = 0

=> - 76 kg m/s + ( 68 kg) (v_Jesse) = 0

=> ( 68 kg) (v_Jesse) = 76 kg m/s

=> v_Jesse =

=> v_Jesse = 1.12 m/s

So, after Jesse’s 1^st throw her speed would be 1.12 m/s away from Joe.

Now, using equation 1 for Joe’s 1^st catch, we have,

( 100 kg) (v_Joe) + ( 2 kg ) (v_asteroid )+ ( 68 kg) (76/68 m/s) = 0

Both Joe and asteroid would be moving with the same speed, v’ after Joe’s 1^st catch.

=> (100 kg ) ( v’) + ( 2 kg) ( v’) + 68 kg m/s = 0

=> ( 102 kg ) (v’) + 68 kg m/s = 0

=> ( 102 kg) (v’) = - 68 kg m/s

=> v’ =

=> v’ = - 0.67 m/s

So, after Joe’s 1^st catch his speed would be 0.67 m/s away from Jesse.

To summarize,

Joe’s speed after his 1^st throw = 0.4 m/s

Jesse’s speed after her 1^st catch = 0.57 m/s

Jesse’s speed after her 1^st throw = 1.12 m/s

Joe’s speed after his 1^st catch = 0.67 m/s