# Jhevon Needs Help With Physics

• Oct 7th 2007, 07:20 PM
Jhevon
Jhevon Needs Help With Physics
My cousin asked me to help her with some of her physics homework questions. It's been ages since i've done physics and even then, the type of questions i did were different from these. i'm currently trying to find ways to attack these problems, but any additional help will be much appreciated.

Please be advised that i posted these questions on 2 other forums. so check if a question you are planning to answer has been answered there

Art of Problem Solving Forum

Help With Physics Homework

Here are the questions:

4.9)

A 0.140 kilogram baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 0.11 meters. What was the average force applied by the ball on the glove?

4.25)

One 3.20 kilogram bucket is hanging by a massless cord from another 3.20-kilogram bucket, also hanging by a massless cord. (a) is the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 meters per second squared by the upper cord, calculate the tension in each cord.

4.54)

A roller coaster reaches the top of the steepest hill with a speed of 6.00 kilometers per hour (1.67 meters per second). It then descends the hill, which is at an average angle of 45.0 degrees and is 45.0 meters long. What is its speed when it reaches the bottom? Assume the kinetic coefficient of friction is 0.18.

5.15)

How many revolutions per minute would a 15-meter-diameter Ferris wheel need to make for its passengers to feel "weightless" at the topmost point?

5.37)

A typical white dwarf star, which was once an average star like our sun but is now in the last stage of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity on this star?

5.48)

During an Apollo lunar landing mission, the common module continued to orbit the moon at an altitude of about 100 kilometers. How long did it take to go around the moon once?

for the first question i tried using the suvat equation $v^2 = u^2 + 2as$ to find the acceleration and then finishing off with F = ma to find the force. but the answer seems weird to me, is this the right way?

for the rest of the questions, i have no clue (at the moment at least)
• Oct 7th 2007, 09:46 PM
earboth
Quote:

Originally Posted by Jhevon
My cousin asked me to help her with some of her physics homework questions. ...
Here are the questions:

4.9)

A 0.140 kilogram baseball traveling 35.0 m/s strikes the catcher's mitt, which, in bringing the ball to rest, recoils backward 0.11 meters. What was the average force applied by the ball on the glove?

...

Hi,

there are (mostly) 3 equations which you need for these kinds of problems:

m = mass
mt = time
d = distance
v = speed
a = acceleration
F = force

The formulae:

$v = a \cdot t$
$d = \frac12 \cdot a \cdot t^2$
$F = m \cdot a$

the ball was braked(?) (negatively accelerated ). You know the speed and the mass, you have to calculate all missing values:

$d = \frac12 \cdot a \cdot t^2~\implies~a = \frac{2d}{t^2}$

Plug in this term into the formula of speed: $v = \frac{2d}{t}$ You know v and d, calculate t:
$35 \ \frac ms= \frac{0.22\ m}{t}~\implies~t\approx 0.006286\ s$

Now calculate a:

$35\ \frac ms = a \cdot 0.006286\ s~\implies ~ a = 5568.2\ \frac{m}{s^2}$

and now calculate the force: $F = 0.14\ kg \cdot 5568.2\ \frac{m}{s^2} \approx 779.5\ N$

(That's a pretty hard hit for a ball travelling at $126\ \frac{km}{h}$ so better check my calculations)
• Oct 7th 2007, 09:50 PM
Jhevon
Quote:

Originally Posted by earboth
Hi,

there are (mostly) 3 equations which you need for these kinds of problems:

m = mass
mt = time
d = distance
v = speed
a = acceleration
F = force

The formulae:

$v = a \cdot t$
$d = \frac12 \cdot a \cdot t^2$
$F = m \cdot a$

the ball was braked(?) (negatively accelerated ). You know the speed and the mass, you have to calculate all missing values:

$d = \frac12 \cdot a \cdot t^2~\implies~a = \frac{2d}{t^2}$

Plug in this term into the formula of speed: $v = \frac{2d}{t}$ You know v and d, calculate t:
$35 \ \frac ms= \frac{0.22\ m}{t}~\implies~t\approx 0.006286\ s$

Now calculate a:

$35\ \frac ms = a \cdot 0.006286\ s~\implies ~ a = 5568.2\ \frac{m}{s^2}$

and now calculate the force: $F = 0.14\ kg \cdot 5568.2\ \frac{m}{s^2} \approx 779.5\ N$

(That's a pretty hard hit for a ball travelling at $126\ \frac{km}{h}$ so better check my calculations)

that's the answer i got using the suvat equations, i was unsure of it myself

i did $v^2 = u^2 + 2as$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement. thus we have:

$0 = (35)^2 + 2(0.11)a$

$\Rightarrow a = 5568.18$ m/(s^2)

thus, $F = ma = (0.14)(5568.18) \approx 779.5$ N

thanks earboth
• Oct 8th 2007, 02:30 AM
topsquark
Quote:

Originally Posted by Jhevon
that's the answer i got using the suvat equations, i was unsure of it myself

i did $v^2 = u^2 + 2as$, where $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration and $s$ is the displacement. thus we have:

$0 = (35)^2 + 2(0.11)a$

$\Rightarrow a = 5568.18$ m/(s^2)

thus, $F = ma = (0.14)(5568.18) \approx 779.5$ N

thanks earboth

Correct, though note your acceleration is negative. It is an artifact of the coordinate system you are using. Since all you are after is the magnitude of the force (questions like this aren't picky enough about their terminology!) you can drop the negative sign.

-Dan
• Oct 8th 2007, 02:43 AM
topsquark
Quote:

Originally Posted by Jhevon
4.25)

One 3.20 kilogram bucket is hanging by a massless cord from another 3.20-kilogram bucket, also hanging by a massless cord. (a) is the buckets are at rest, what is the tension in each cord? (b) If the two buckets are pulled upward with an acceleration of 1.60 meters per second squared by the upper cord, calculate the tension in each cord.

Do you know what a "Free Body Diagram" is? You want to "sketch" each bucket in their own diagram (a point is all you need to represent the bucket) and label all the forces on that bucket.

So we have two buckets here. The FBD for the bottom bucket has two forces on it: a tension $T_{b1}$ acting upward and a weight $w_b$ acting downward. The FBD for the top bucket has a weight $w_t$ acting downward, a tension in the top string $T_{t2}$ acting upward, and a tension in the bottom string $T_{t1}$ acting downward.

These diagrams apply for both cases, where we have no acceleration and where we have acceleration.

So for the no acceleration case.

I am defining a +y direction to be upward in both FBDs.

Apply Newton's 2nd Law to the bottom bucket.
$\sum F_b = T_{b1} - w_b = m_b a_b$

Here the acceleration of the bottom bucket is 0 m/s^2, so
$T_{b1} - w_b = 0$

Thus the tension in the bottom string is equal to the magnitude of the weight of the bottom bucket.

Now apply Newton's 2nd to the top bucket:
$\sum F_b = T_{t2} - T_{t1} - w_t = m_t a_t$

Again $a_t = 0~m/s^2$ so we have:
$T_{t2} - T_{t1} - w_t = 0$

But what is $T_{t1}$? It is the tension in the bottom string. Since the strings are ideal (that is to say massless and inextensible) the tension in the string is a constant along the string. So $T_{t2} = T_{b1}$ (in magnitude.) Thus
$T_{t2} - w_b - w_t = 0$

$T_{t2} = w_b + w_t$

This makes sense qualitatively since the top string is supporting both buckets.

The acceleration case is the same, it's just that the accelerations in the Newton's Law equation is not zero anymore, so the tensions in the strings are going to be larger, which you can verify.

-Dan
• Oct 8th 2007, 03:09 AM
topsquark
Quote:

Originally Posted by Jhevon
4.54)

A roller coaster reaches the top of the steepest hill with a speed of 6.00 kilometers per hour (1.67 meters per second). It then descends the hill, which is at an average angle of 45.0 degrees and is 45.0 meters long. What is its speed when it reaches the bottom? Assume the kinetic coefficient of friction is 0.18.

This is a "Work-Energy" problem.

We have a kinetic friction force acting directly opposite the infinitesimal displacement. (Which is always the case.) Since we are dealing with (on average) a straight line motion we have that the kinetic friction force is directly opposite the overall displacement. And kinetic friction is a non-conservative force.

So, the work-energy theorem says:
$W_{nc} = \Delta E$

$\vec{f_k} \cdot \vec{s} = E - E_0$
( $\vec{s}$ is the displacement.)

Since the friction is in the oppostite direction of the
displacement:
$-f_k s = E - E_0$

Now, E is the total energy. That is,
$E = KE + PE$

We need to set a 0 point level for the potential energy, and I'm going to choose that to be at the bottom of the "hill." The positive direction is always against the direction of the weight, so it is straight upward.

Thus
$-f_k s = (KE + PE) - (KE_0 + PE_0)$

$-f_k s = \left ( \frac{1}{2}mv^2 + mgh \right ) - \left ( \frac{1}{2}mv_0^2 + mgh_0 \right )$

We know $v_0$ and we are looking for $v$. We have defined a 0 for our PE to be at the bottom of the hill, so $h = 0~m$. What is $h_0$?

Well, we have an 45 m long incline at 45 degrees, so
$h = 45sin(45^o)$.
(I'm ignoring my typical anal compulsion to include units for the sake of simplicity.)

Thus
$-f_k s = \left ( \frac{1}{2}m \cdot 1.67^2 \right ) - \left ( \frac{1}{2}mv_0^2 + mg(45)sin(45^o) \right )$

Now what? Well $f_k = \mu _k N$, so we need to know the magnitude of the normal force on the car.

FBD time again. I'm going to choose a coordinate system where +x is in the direction of the car down the slope and +y is perpendicular to this, upward from the track. We have 3 forces in the FBD: a weight w acting directly downward, a kinetic friction force $f_k$ acting in the -x direction, and a normal force N acting in the +y direction.

Since all we need is the magnitude of N all I'm going to worry about is the y direction. So:
$\sum F_y = N - w_y = ma_y$

Since there is no acceleration in the y direction (else the car would either lift off the track or be pushed down into the track, both illogical statements):
$N - w_y = 0$

$N = w_y = w~cos(45^o) = mg~cos(45^o)$

So the friction force becomes:
$f_k = \mu _k mg~cos(45^o)$

and the work-energy formula becomes:
$- \mu _k mg~cos(45^o) s = \left ( \frac{1}{2}m \cdot 1.67^2 \right ) - \left ( \frac{1}{2}mv_0^2 + mg(45)sin(45^o) \right )$

Note that we have a factor of m in all terms, so this divides out. (Which is good since we were never given it in the first place. :) )
$- (0.18) g~cos(45^o) (45) = \left ( \frac{1}{2} \cdot 1.67^2 \right ) - \left ( \frac{1}{2}v_0^2 + g(45)sin(45^o) \right )$

You can finish from here.

-Dan
• Oct 8th 2007, 03:19 AM
topsquark
Quote:

Originally Posted by Jhevon
5.15)

How many revolutions per minute would a 15-meter-diameter Ferris wheel need to make for its passengers to feel "weightless" at the topmost point?

A feeling of weightlessness is caused by the "apparent weight" being equal to 0 N. The apparent weight is simply another term for the normal force.

So, the net force on any object moving in a circle at constant speed is equal to the centripetal force. So I'm going to consider the rider to be at the top of the Ferris wheel and do a Newton's 2nd problem on it at that point.

FBD: I'm choosing a +y direction straight up. (We don't need a +x here, so I'll ignore it.) There is a weight w on the rider acting straight down and a normal force N from the seat acting straight up. The centripetal force is a net force so it does not belong in this diagram!!

Newton's 2nd says:
$\sum F_y = N - w = F_c$

Now, the centripetal force has a magnitude of
$F_c = \frac{mv^2}{r}$
in the direction of the center of the circle. Thus it is acting downward, so
$N - w = -F_c = -\frac{mv^2}{r}$

Now, the person is "weightless." Thus N = 0 N:
$-w = -\frac{mv^2}{r}$

Or
$v = \sqrt{\frac{wr}{m}} = \sqrt{\frac{mgr}{m}} = \sqrt{gr}$

Now we have the translational speed of the Ferris wheel, we need to convert this to an angular speed. Well
$v = r \omega$
where $\omega$ is the angular speed. Thus
$\omega = \frac{v}{r} = \frac{\sqrt{gr}}{r} = \sqrt{\frac{g}{r}}$.

I'll let you convert the angular speed in rad/s to rev/min.

-Dan
• Oct 8th 2007, 03:23 AM
topsquark
Quote:

Originally Posted by Jhevon
5.37)

A typical white dwarf star, which was once an average star like our sun but is now in the last stage of its evolution, is the size of our moon but has the mass of our sun. What is the surface gravity on this star?

"Surface gravity" is, contrary to what you might think, is an acceleration, not a force. It is the acceleration due to an objects mass (gravity) at the surface of the object.

So
$g = \frac{GM}{r^2}$
where M is the mass of the object and r is the radius of the (spherical, in this case) object.

So look up the solar mass and Moon's radius and you'll have it. (I'm a theorist so I don't care about the actual numbers. (Wink) )

-Dan
• Oct 8th 2007, 03:33 AM
topsquark
Quote:

Originally Posted by Jhevon
5.48)

During an Apollo lunar landing mission, the common module continued to orbit the moon at an altitude of about 100 kilometers. How long did it take to go around the moon once?

If we assume a circular orbit then the net force on the module is equal to the centripetal force. When you do the FBD you find there is only one force acting on the module: the gravity from the Moon. (Well, there's one from the Earth too (and the Sun, and Jupiter, and the Andromeda galaxy for that matter), but this is too small to worry about.) So the centripetal force is equal to the force of gravity on the module:
$F_c = F_G$

$\frac{mv^2}{R} = \frac{GmM}{R^2}$
where m is the mass of the module, M is the mass of the Moon, and R is the distance from the center of the Moon to the module, NOT the height of the module above the Moon! So let's call $R = r + h$ where r is the radius of the Moon and h is the height (altitude) above the Moon.

So
$v = \sqrt{\frac{GM}{r + h}}$
(after canceling some pesky factors)

We want the period of the motion. Well, we have the translational speed, and this is related to the angular speed by:
$\omega = \frac{v}{R} = \frac{\sqrt{\frac{GM}{r + h}}}{r + h} = \sqrt{\frac{GM}{(r + h)^3}}$

and $\omega$ is related to the period T by
$T = \frac{2 \pi}{\omega} = 2 \pi \sqrt{\frac{(r + h)^3}{GM}}$
which you could have conceivably looked up from the start. (I never remember such formulas, only their derivations.)

-Dan
• Oct 8th 2007, 08:33 AM
Jhevon
Quote:

Originally Posted by topsquark
Correct, though note your acceleration is negative. It is an artifact of the coordinate system you are using. Since all you are after is the magnitude of the force (questions like this aren't picky enough about their terminology!) you can drop the negative sign.

-Dan

i realized that, but i felt weird writing a negative sign in front of the force for some reason, so i just dropped the negatives all together
• Oct 8th 2007, 08:35 AM
Jhevon
Thanks a lot topsquark!!! You are the man!!! (Bow)
• Oct 8th 2007, 08:47 AM
ThePerfectHacker
I just want to give a tip. If you post homework problems on AoPs they will probably not be answered.
• Oct 8th 2007, 08:48 AM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
I just want to give a tip. If you post homework problems on AoPs they will probably not be answered.

ah, ok. i didn't know that. i think they have a homework section on there though
• Oct 8th 2007, 01:36 PM
topsquark
Quote:

Originally Posted by Jhevon
Thanks a lot topsquark!!! You are the man!!! (Bow)

Well, since Krizalid won't take that title... :)

-Dan
• Oct 8th 2007, 01:42 PM
Jhevon
Quote:

Originally Posted by topsquark
Well, since Krizalid won't take that title... :)

-Dan

yeah, lol, he's not "the man" he's not "the woman" we should just call him "the" ...or perhaps more appropriately, "el"