Question

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange. They decide to trade. Atronaut 1 tosses the 0.125 kg apple toward astronaut 2 with a speed of Vi1=1.13 m/s. The 0.160 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.18 m/s. Unfortunately, the fruits collide, sending the orange off with a speed of 0.999 m/s in the negative Y direction.

A) What are the final speed of the apple in this case (Vf1)?

B) What are the direction of the apple in this case (theta)?

Answer 1

Explanation:

Mass of Apple, $m_a=0.125\ kg$

Mass of orange, $m_o=0.16\ kg$

Initial speed of orange, $u_o=1.18i$

Initial speed of Apple, $u_a=-1.13i$

Final speed of the orange, $v_o=-0.999j$

(a) Let $v_a$ is the final sped of the apple. It can be calculated using the law of conservation of momentum as :

$m_au_a+m_ou_o=m_av_a+m_ov_o$

$m_av_a=(m_au_a+m_ou_o)-m_ov_o$

$m_av_a=(0.125\times (-1.13i)+0.16\times (1.18i))-0.16\times (-0.999j)$

$m_av_a=(-0.141i+0.1888i)+0.159j$

$m_av_a=0.0478i+0.159j$

$v_a=\dfrac{0.0478i+0.159j}{0.125}$

$v_a=0.382i+1.27j$

$|v_a|=\sqrt{0.382^2+1.27^2}=1.32\ m/s$

So, the final speed of the apple is 1.32 m/s

(b) The direction of the apple in this case is given by:

$tan\theta=\dfrac{1.27}{0.382}$

$\theta=73.25^{\circ}$

Hence, this is the required solution.