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°.

Answer 2

The speed of the apple tossed by the astronauts is 1.448 m/s and the direction of this apple is 66.65 degrees.

What is conservation of momentum?

Momentum of a object is the force of speed of it in motion. Momentum of a moving body is the product of mass times velocity.

When the two objects collides, then the initial collision of the two body is equal to the final collision of two bodies by the law of conservation of momentum.

Two astronauts on opposite ends of a spaceship are comparing lunches. One has an apple, the other has an orange.

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 .

The equation for the momentum in x direction can be given as,

$m_{a}\times v_{xa}+0=m_pv_p-m_ov_o$

Here, subscript a and o represent the apple and orange, respectively.

Put the values,

$0.110\times v_{xa}+0=0.110(1.13)-0.150(1.25)\\v_{xa}=-0.574\rm m/s$

The fruits collide, sending the orange off with a speed of 0.977 m/s in the negative y direction.

So the velocity in y-direction,

$m_{a}\times v_{ya}-m_{o}\times v_{o}=0\\v_{ay}=1.33\rm m/s$

So the speed of the apple is,

$V=\sqrt{v_{xa}^2+v_{ya}^2}\\V=\sqrt{(-0.574)^2+1.33^2}\\V=1.448\rm m/s$

The direction of the apple is,

$\tan\theta=\dfrac{v_{ya}}{v_{xa}}\\\tan\theta=\dfrac{1.33}{0.574}\\\theta=66.65^o$

Thus, the speed of the apple tossed by the astronauts is 1.448 m/s and the direction of this apple is 66.65 degrees.

Learn more about the conservation of momentum here;

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