Physics problems Classical Mechanics and Gravitation

[Superdreamer]

Hi i was wondering if anyone could help me with the following problems please,I'm a bit confused!

Q1. A 9.4 kg projectile is fired vertically upwards. Air drag dissipates
68 kJ during its ascent. How much higher would it have gone were air drag
negligible?

Here I used the formula mgh and worked out the height to be 738.17m.Is this right If not what formula and method should I use?

q2 Suppose the Sun were to run out of nuclear fuel and collapse to form a
white dwarf star, with a radius equal to that of the Earth (6.4 x 10 6 m).
Assuming no mass loss, what would then be the Sun’s new rotational
period. Assume that both the Sun and the white dwarf are uniform solid
spheres. (The period of the sun is currently about 25 days. The moment of
inertia of a sphere is 5
r M 2 2
and the solar radius is 6.7 x 10 8 m.)

Completely lost on this q don't know where to start?

Q3 The Martian satellite Phobos travels in an approximately circular orbit of
radius r = 9.4 x 10 6 m with a period T of 459 minutes. Calculate the mass of
Mars from this information.

I used keplers 3rd law here Gm=4piR^2/T^2 and found the mass to be 6.43 x10^-13 is this the correct way for doing this?

 

[Andrew Mason]

Hi i was wondering if anyone could help me with the following problems please,I'm a bit confused!

Q1. A 9.4 kg projectile is fired vertically upwards. Air drag dissipates
68 kJ during its ascent. How much higher would it have gone were air drag
negligible?

Here I used the formula mgh and worked out the height to be 738.17m.Is this right If not what formula and method should I use?

Your answer is correct.

q2 Suppose the Sun were to run out of nuclear fuel and collapse to form a white dwarf star, with a radius equal to that of the Earth (6.4 x 10 6 m).
Assuming no mass loss, what would then be the Sun’s new rotational
period. Assume that both the Sun and the white dwarf are uniform solid
spheres. (The period of the sun is currently about 25 days. The moment of
inertia of a sphere is 5
r M 2 2
and the solar radius is 6.7 x 10 8 m.)

Angular momentum must be conserved in this scenario:
$$L = I\omega = L' = I'\omega'$$
where:
[tex]I = \frac{2mr^2}{5}[/itex]

Since m remains constant, $\omega$ must change as r changes. Work it out.

Q3 The Martian satellite Phobos travels in an approximately circular orbit of radius r = 9.4 x 10 6 m with a period T of 459 minutes. Calculate the mass of Mars from this information.

I used keplers 3rd law here Gm=4piR^2/T^2 and found the mass to be 6.43 x10^-13 is this the correct way for doing this?

I don't think you have Kepler's third law stated correctly.

The centripetal acceleration has to be supplied by gravity. The centripetal acceleration for Phobos is: $a_c = \omega^2 r$. The gravitational acceleration is:

$$a_g = \frac{GM_{mars}}{r^2}$$

Equate the two and work out $M_{mars}$ from that.

AM

 

[quicknote]

Q1. To calculate the height the projectile would have reached without air drag, we need to use the conservation of energy principle. The initial energy of the projectile is equal to its final energy, which is given by the potential energy at the maximum height. So, we can write the equation as follows:

Initial energy = Final energy

Initial energy = Kinetic energy + Potential energy

Final energy = Potential energy

Therefore, we can write the equation as:

mgh = mgh + 68 kJ

Where m is the mass of the projectile, g is the acceleration due to gravity, and h is the height reached by the projectile without air drag. Rearranging the equation, we get:

h = (mgh - 68 kJ)/mg

h = (9.4 kg x 9.8 m/s^2 x h - 68,000 J)/9.4 kg x 9.8 m/s^2

h = 738.17 m

So, your answer is correct.

Q2. To calculate the new rotational period of the Sun, we need to use the conservation of angular momentum principle. The initial angular momentum of the Sun is equal to its final angular momentum, which is given by the moment of inertia and the angular velocity. So, we can write the equation as follows:

Initial angular momentum = Final angular momentum

Initial angular momentum = Moment of inertia x Angular velocity

Final angular momentum = Moment of inertia x Angular velocity

Therefore, we can write the equation as:

I1ω1 = I2ω2

Where I is the moment of inertia, ω is the angular velocity, and the subscripts 1 and 2 represent the initial and final values, respectively. We can also write the equation as:

I1ω1 = I2ω2

I1(2π/T1) = I2(2π/T2)

Where T is the rotational period. Substituting the values given in the question, we get:

(5/6)(6.7 x 10^8 m)^2 x (2π/25 days) = (5/6)(6.4 x 10^6 m)^2 x (2π/T2)

Solving for T2, we get:

T2 = 3.24 hours

So, the new rotational period of the Sun would be 3.24 hours.

Q3. Yes, you