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**1. The problem statement, all variables and given/known data**

A NASA satellite has just observed an asteroid that is on a collision course with Earth. The asteroid has an estimated mass, based on its size, of

**5 x 109 kg**. It is approaching the Earth on a head-on course with a velocity of

**600 m/s**relative to the Earth and is now

**5 x 106 km**away.

M

_{asteroid}= 5 x 10^9 kg

M

_{earth}= 5.98 x 10^24 kg

V

_{i}= 600 m/s

r

_{i}= 5 x 10^9 m

a. With what speed will it hit the Earth’s surface?

b. How much thermal energy will be released?

c. How much will the impact change the orbital velocity of the Earth?

**2. Relevant equations**

K

_{i}+ U

_{i}= K

_{f}+ U

_{f}

½M

_{A}V

_{i}² - G(M

_{E}M

_{A})/r

_{i}= ½M

_{A}V

_{f}² - G(M

_{E}M

_{A})/r

_{E}

½mv

_{f}² - U

_{g}= ½mv² - U

_{g}

M

_{A}V

_{A}= (M

_{A}+ M

_{E})V

M

_{A}V

_{A}+ M

_{E}V

_{E}= (M

_{A}+ M

_{E})V

_{f}

V

_{f}= (M

_{A}V

_{A}) / (M

_{A}+ M

_{E})

**3. The attempt at a solution**

Part A we worked out using the first two equations in our study group:

*½ (5 x 10^9 kg)(600 m/s)² - G(5.98x10^24 kg)( 5 x 10^9 kg)/(5 x 10^9 m)*

and we ended up with

**1.12 x 10^4 m/s**relative to Earth as our final answer.

Part B we were unsure how to work because we haven't covered anything in class regarding how much energy is released as thermal energy, light, and sound. Using the third and fourth equations from above we worked it out as such:

*½(5 x 10^9 kg)(1.12 10^4 m/s)² - ½(5 x 10^9 kg)(5.017 x 10^-13)²*

and ended up with a final answer of

**3.136 x 10^17 J**.

Part C also left me with a certain level of uncertainty. Using the last two equations listed, I ended up with

**9.37 x 10^-12 m/s**or

**3.14 x 10^15 %**

**1. The problem statement, all variables and given/known data**

Two trees are

**7.6 m**apart. Calculate the magnitude of the force F a backpacker must exert to hold a

**16 kg**backpack so that the rope sags at its midpoint by:

a. 1.5 m

b. 0.15 m

**2. Relevant equations**

Σt = (TsinΘ)(x) - mgh = 0

**3. The attempt at a solution**

For Part A we were confused as to what values we should use for x and h because the diagram shows the rope going across the trees to a pulley where the camper is applying the force to the rope. Some of us thought that we should use a distance of

*8.18 m*for x and

*4.09 m*for h based on the numbers we came up using trig. However, some argued that we should use distances of

*7.6 m*and

*3.8 m*because that is the distance between the trees. I maintained the opinion that the pulley would create more length in the rope, so I went with the first set of numbers I mentioned.

Θ = arctan(1.5/3.8) = 21.54º

Σt = Tsin(21.54º)(8.18 m) - (16 kg)(9.8 m/s²)(4.09 m) = 0

Θ = arctan(1.5/3.8) = 21.54º

Σt = Tsin(21.54º)(8.18 m) - (16 kg)(9.8 m/s²)(4.09 m) = 0

My final answer for part A was

**213.54 N**

Part B offered no confusion because the distances obtained using trig were essentially

*7.6 m*and

*3.8 m*