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. Astronaut 1 tosses the 0.110 kg apple toward astronaut 2 with a speed of vi,1 = 1.13 m/s . The 0.150 kg orange is tossed from astronaut 2 to astronaut 1 with a speed of 1.25 m/s . Unfortunately, the fruits collide, sending the orange off with a speed of 0.977 m/s in the negative y direction.

Answer 1

Answer:

The speed and direction of the apple is 1.448 m/s and 66.65°.

Explanation:

Given that,

Mass of apple = 0.110 kg

Speed = 1.13 m/s

Mass of orange = 0.150 kg

Speed = 1.25 m/s

Suppose we find the final speed and direction of the apple in this case

Using conservation of momentum:

Before:

In x direction,

$P_{b}=m_{p}v_{p}-m_{o}v_{o}$

$P_{b}=0.110\times1.13-0.150\times1.25$

$P_{b}=−0.0632\ kg-m/s$

In y direction = 0

After:

$v_{ay}$ is velocity of the apple in the y direction

$v_{ax}$ is the velocity of the apple in the x direction

Momentum again:

In x direction,

$0.110\times v_{ax}+0=−0.0632$

$v_{x}=\dfrac{−0.0632}{0.110}$

$v_{x}=−0.574\ m/s$

In y-direction,

$0.110\times v_{ay}-0.150\times0.977=0$

$v_{ay}=\dfrac{0.150\times0.977}{0.110}$

$v_{ay}=1.33\ m/s$

We need to calculate the speed of apple

$v_{a}=\sqrt{(v_{x})^2+(v_{y})^2}$

Put the value into the formula

$v_{a}=\sqrt{(−0.574)^2+(1.33)^2}$

$v_{a}=1.448\ m/s$

We need to calculate the direction of the apple

Using formula of angle

$\tan\theta=\dfrac{v_{ay}}{v_{ax}}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{1.33}{0.574})$

$\theta=66.65^{\circ}$

Hence, The speed and direction of the apple is 1.448 m/s and 66.65°.