# AP Physics Homework (Chapter 7) [7-1 ~ 7-2]

October 24th, 2009 by ben
7-1 and 7-2 Momentum and Its Conservation

1. (I) What is the magnitude of the momentum of a 28-g sparrow flying with a speed of 8.4 m/s?

$\mathbf{\bar{p}} = m \hat{v}$

Trivial.

$\mathbf{\bar{p}} = 0.028 \cdot 8.4 = \boxed{0.2352\, \dfrac{\mathrm{kg\,m}}{\mathrm{s}^2}}$

2. (I) A constant friction force of 25 N acts on a 65-kg skier for 20 s. What is the skier’s change in velocity?

Change in velocity = acceleration. We could do this with kinematics ($F_{\mathrm{net}} = ma$ and then using $a$ with kinematics equations):

$25 = 65 \cdot a$

Thus the acceleration is:

$a = 0.384615 \dfrac{\mathrm{m}}{\mathrm{s}^2}$

And by accelerating at this rate for 20 seconds, the change in velocity is:

$\Delta v =0.384615 \cdot 20 = \boxed{7.69\, \mathrm{m/s}}$

However, using impulse (change in momentum):

$\Delta p = F \cdot \Delta t$

Plugging in (we do this too much in physics):

$\Delta p = 25 \cdot 20 = 500\, \mathrm{kg\, m/s}$

And since momentum equals mass times velocity, the change in velocity is 500 divided by the mass, 65 kg.

$\Delta v = 7.69\, \mathrm{m/s}$

Oh, but don’t forget, we’re decelerating, so the change is negative (velocity is decreasing).

$\Delta v = \boxed{-7.69\, \mathrm{m/s}}$

*skipskip*

5. (II) Calculate the force exerted on a rocket, given that the propelling gases are expelled at a rate of 1500 kg/s with a speed of $4.0 \times 10^4 m/s$ (at the moment of takeoff).

Choose $\Delta t$ to be one second.

$F = \dfrac{\Delta p}{\Delta t} = \dfrac{1500 \cdot 4.0 \times 10^4}{1} = 6000 \times 10^4 = \boxed{6.0 \times 10^7\, \mathrm{N}}$

Such trivial questions.

11. (II) An atomic nucleus initially moving at 420 m/s emits an alpha particle in the direction of its velocity, and the remaining nucleus slows to 350 m/s. If the alpha particle has a mass of 4.0 u and the original mucleus has a mass of 222 u, what speed does the alpha particle have when it is emitted?

Conservation of momentum.

$420 \cdot 222 = s \cdot 4 + 350 \cdot 218$

I don’t even want to bother solving this, it’s so trivial.

$s = \dfrac{420 \cdot 222 - 350 \cdot 218}{4} = \boxed{4235 \,\mathrm{m/s}}$

*skip*

13. (III) A 975-kg two-stage rocket is travelling at a speed of $5.80 \times 10^3 \,m/s$ with respect to Earth when a pre-designed explosion separates the rocket into two sections of equal mass that then move at a speed of $2.20 \times 10^3 \,m/s$ relative to each other along the original line of motion.

(a) What are the speed and direction of each section (relative to Earth) after the explosion?

(b) How much energy was supplied by the explosion? [Hint: what is the change in KE as a result of the explosion?

It looks long, but I bet this one’s trivial too.

(a) Hm, it looks like the main hard part is how the speed is given to you “relative to the other rocket section”.

It just means that speed 1 minus speed 2 = the speed “relative to the other” I guess, since the directions of the velocities are all still the same.

$m_i v_i = m v_1 + m v_2$

And substituting the “$2.20 \times 10^3 \,m/s$ relative to each other”:

$m_i v_i = m v_1 + m (v_1 + 2.20 \times 10^3)$

Plugging in the given variables:

$(975) (5.80 \times 10^3) = \left(\dfrac{975}{2}\right) \left( v_1 \right) + \left(\dfrac{975}{2}\right) \left( v_1 + 2.20 \times 10^3 \right)$

$v_1 + v_1 + 2.20 \times 10^3 = 11600$

$2v_1 = 9400$

$v_1 = \boxed{4700\, \mathrm{m/s}}$

And adding 2200 will give us the velocity of the faster rocket stage.

$v_2 = \boxed{6900\, \mathrm{m/s}}$

(b) We’re asked to find the change in KE (Kinetic Energy) from the explosion separating the rocket stages. All we have to do basically is subtract the final KE from the initial. Trivial.

$KE_i - KE_f = \Delta KE$

Since the formula for KE is $\frac{1}{2}mv^2$, all this problem is is plugging in. Like every other physics problem.

$\frac{1}{2} (975) (5800)^2 - \frac{1}{2} (975/2) (4700^2 + 6900^2) = \Delta KE$

Evaluating is always such a bother. Ugh.

$\frac{1}{2} (975) (5800^2 - \frac{1}{2}(4700^2 + 6900^2)) = \Delta KE$

The KE change from final to initial (I now realize I shoulda flipped it the other way, hehe) is:

$\Delta KE = -5.899 \times 10^8$

Flipping the sign, the kinetic energy produced by the explosion is:

$\Delta KE = \boxed{5.899 \times 10^8 \,\mathrm{J}}$

14. (III) A rocket of total mass 3180 kg is traveling in outer space with a velocity of 115 m/s. To alter its course by $35.0^\circ$, its rockets can be fired briefly in a direction perpendicular to its original motion. If the rocket gases are expelled at a speed of 1750 m/s, how much mass must be expelled?

[spoiler]
I think this would be easier with forces, so I’ll do it with forces first before doing it with conservation of momentum or whatever. The original force plus the force caused by the perpendicular expulsion of rocket gases must have an angle of 35 degrees.

$\tan \left(\dfrac{F_R}{F_G} \right) = 35^\circ$

Where R represents the rocket and G represents the expelled gases. Problem: we don’t know the acceleration. Nevermind then, let’s use momentum. Same deal, whatever.

$\tan \left(\dfrac{p_R}{p_G} \right) = 35^\circ$

$\dfrac{3180 \cdot 115}{m \cdot 1750} = \arctan 35^\circ$

$m = \dfrac{1.54233 \cdot 1750}{3180 \cdot 115}$
[/spoiler]

Damn, I did this wrong. Maybe it wasn’t so trivial after all.

I feel stupid now.

$v_P = v_R \tan 35^\circ$

[spoiler]
Where “P” indicates the vector in the direction of propulsion, and “G” is the same vector in the opposite direction (the velocity of the gas itself). Since, by conservation of momentum:

$p_i = p_f \to m_R v_R = m_G v_G + (m_R - m_G) v_P$

Since $v_G$ is just $-v_P$, we can substitute:

$p_i = p_f \to m_R v_R = - m_G v_P + (m_R - m_G) v_P$

Now, we’re trying to find $m_G$, the mass of gas that needs to be expelled.

$-2 m_G v_P = m_R v_R - m_R v_P$

$m_G = \dfrac{m_R (v_R - v_P) }{ -2 v_P }$

$m_G = \dfrac{m_R v_R (1 - \tan 35^\circ)}{ -2 ( v_R \tan 35^\circ ) }$
[/spoiler]

WOOOW. My second approach was wrong too. Ugh, lol.

Following along with the textbook’s approach:

$p_i = p_i \to 0 = m_G v_G + (m_R - m_G) v_P$

Basically, instead of comparing, like, the total momentum, they’re only comparing the momentum in the perpendicular axis. That’s smart…

$m_G = \dfrac{m_R v_P}{v_P - v_G}$

Thus.

$m_G = \dfrac{(3180)(115 \tan 35^\circ)}{(115 \tan 35^\circ) - (-1750)} = \boxed{14.0 \,\mathrm{kg}}$

### Related Posts:

1. Chantal says:

oh gosh you are brilliant. and darn good at math!
thanks so much ;)

2. Benji says:

3. Nook says:

What’s the answer to number 14??

4. Benji says:

it’s the last problem on the post. answer is 14.0 kg.

5. Hello, this is really a actually fascinating internet blog and ive loved reading several on the articles and posts contained upon the web site, sustain the terrific perform and hope to go through a lot more thrilling articles or blog posts within the time to come.

6. Celina says:

Hi Benji.

Just so you know it is stated it is supposed to relative to the earth :D just if you would like to change it :DÂ

7. dude says:

god damn, they’ve been using the same questions for so long… which is a very good thing. and thanks 9 years late!!!

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