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I found something that discussed the relationship between addition of probabilities and number of ways. Here is what it says: Suppose the event can happen in two ways which cannot concur and let ##\frac {a_1}{b_1}##, ##\frac {a_2}{b_2}## be the chances of the happening of the event in these...

##\displaystyle P(\text{white})=P(\text{box 1 AND white})\,\text{OR}\,P(\text{box 2 AND white})=\frac{1}{2}\cdot\frac{20}{40}+\frac{1}{2}\cdot\frac{10}{15}=\frac{7}{12}## That is the correct answer from the other thread. From the denominator of this final result, we can conclude that the...

But, this is the definition of probability itself. So, isn't that should be true for any cases?

https://www.physicsforums.com/threads/choosing-a-ball-at-random-from-a-randomly-selected-box.1034377/ First of all, I would like to point out that this is the same exact question from what is being discussed in the thread above. In that thread, the problem is solved by adding the probability...

So, to find ##k##, should I make use of the quotient instead of the remainder? The quotient will be : ##\frac n k + \frac {k-1} {k²} ## Here is what I thought initially: At first glance, the only possible value of ##k## is 1. Otherwise, the second term will become a fraction (consequently...

So, ##n\, |\, (p − 1)## implies ##p = nk + 1## and ##p ≥ n + 1##. Clearly, ## p \,|\, n^3 − 1## implies either : ##p \,|\, n − 1 ## (which is impossible, because p cannot be less than ##n-1##) or ##p \,|\,n^2 + n + 1##. Now, our main focus is ##p\, |\,n^2 + n + 1##. Since ##p = nk + 1##...

Why can't we just substitute ##(1+x)^{1/x}## as ##e?## then, this equation will become: $$\lim _{x\to 0} \frac{f(x)-f(0)}{x} = \lim _{x\to 0} \frac {e- e}{x}$$ Since x is near zero, then we can make it like this: $$\lim _{x\to 0} \frac{f(x)-f(0)}{x} = \lim _{x\to 0} \frac{0}{x}$$ So, the...

Ah. I see.. Now I managed to prove it. There are some things that is bothering me,though.. 1. Why my method in post 1 does not work? 2. I tried a different way to transform the term ##ln(1+x)##, but I only get the first term of the maclaurin series. Here is what I did: let ##y= ln(1+x)##...

Ok. so, I tried to differentiate the ##(1+x)^{1/x}##, and using the rule, I get: $$\lim _{x\to 0} \frac{x}{(1+x)^{1/x} - e} = \frac{1}{\frac{x(1+x)^{-1+\frac{1}{x}}\,\,-\,\,(1+x)^{\frac 1 x}\,\,ln(1+x)}{x^2}}$$ Then, all I need to do is to substitute the value of ##x## as zero, right? The...

Right now, I am trying to prove this : I tried to use this identity to solve it: Then, the limit will become ##\frac {x}{e-e}## However, the result is still ##\frac 0 0 ## Could you please give me hints to solve this problem?

I have a different way in solving the problem, but strangely, the result is different from that written in the solution manual. My method: Firstly, we will solve the ##AB=A## equation $$AB=A$$ $$B=A^{−1}A$$ $$B=I$$ where ## I## is an identity matrix Similarly, we can solve ##BA=B## using the...

Umm... well, it is is true, but what is the relation of it with the area of ring? Why, though? We can't use ##ds## as its height since it is not perpendicular to the bases.

Yes. But, ##ds## is not the height of the trapezoid. So, in my opinion, we are not supposed to multiply this with the ##2πa sin\theta## to get the area.

Why the area of the thin rings are ##2πasin\theta \, ds##? (a is the radius of the hollow sphere) If we look from a little bit different way, the ring can be viewed as a thin trapezoid that has the same base length ( ##2πa sin\theta##), and the legs are ## ds##. The angle between the leg and...

$$\int -mg \mu d(l) = \frac 1 2 m(v²-u²)$$ Where v is the speed of the lower block just before the 2nd collision. $$v = \sqrt {u² - 2g\mu l_0}$$ So, there are two answers: 1. ## v = 0## if ##u² ≤2g\mu l_0## 2. ## v = \sqrt {u² - 2g\mu l_0}## if ##u² ≥2g\mu l_0## Do I also need to think of...

But, In my thread about a rocket ejecting mass, which you can find it here: https://www.physicsforums.com/threads/a-rocket-ejecting-mass.1044968/ we do not ignore the small change in speed (##dv##), even though the force imparted to the rocket (by ejecting the mass) is small. This is why I...

When the upper block collides with the wall, the impulse to it will be: $$ \int (-N+ f)dt = -2mu$$ Where N is the contact force from the wall. Since N is really huge, the impulse from friction in the first equation above can be ignored, hence the equation become: $$ \int -Ndt = -2mu$$...

How can we know that it is so big? The force that is exerted by the wedge to the disc might be 10 N, 50N, etc.

Now I'm confused... When do we can or cannot ignore an impulse? For example, Let's say that I release a ball from a building, and after that it will hit the ground. At the instant when it collides with the ground, can we ignore the impulse by gravity?why?

$$\frac {u²} {8gsin\theta} (2\sqrt {1+sin²\theta} - cos\theta)$$

The wedge's speed always increases, while the horizontal speed of the disc always decreases. If the horizontal speed of the wedge is larger than the disc's, then the disc will start to slide down. We can use the wedge's frame of reference to prove this. If it is larger, then the disc's...

So, when the mass reached the peak, its horizontal velocity will be the same as the wedge's. Using conservation of momentum : $$ mu = 2mv$$ $$v = \frac u 2$$ With v is the final velocity for both objects. Now, what we need is the acceleration of the wedge, which we can find by using Newton's...

I have a difficulty in understanding the question. Fictitious force is a force whose motion is described using a non-inertial frame of reference. Which frame is the question referring to?

The second method in my first post is what is written in the book.(By the way the sentence that I just quoted is also taken from it). Then, I tried a slightly different way by regarding ##dm## as a positive mass, but the result is different from the book's. That is what I'm confused right now...

If I solve it like usual, the result (or, we can also say the area under the curve) is negative. So, does that mean I need to multiple the result by (-1) so that it become positive? EDIT: nevermind. This is clearly incorrect.

"during a small time dt, a negative mass dm gets added to the rocket, and a positive mass (−dm) gets shot out the back"

ah... Um... imo, the dx need to be positive in order to solve it like usual. What should I do ?

I copied it from my textbook. I doubt that its answer is wrong

I have a question. If we assume that ##dm## is positive, is the answer supposed to be different from the one when we regard the ##dm## as negative? 1. If I assume that ##dm## is positive: By using momentum conservation, we will get $$mv=(m-dm)(v+dv)+dm (v-u)$$ simplify the equation $$m \,dv=dm...

They are free to rotate. The total horizontal velocity in lab frame for the upper/lower mass is ##V_{rotation} \, sin \phi + V_{translation\,in\, x \,direction}##. But, is the translation velocity always the same with the middle mass's?

If initially the system is in a straight line (##\phi = 90 ^\circ##) and at rest, then we struck the mass at the center so that it has velocity ##v_o## to the right, after some time (when the velocity of the mass at the center is ##v##), will the upper and lower masses' translation velocities...

Now I understand. Thank you, all of you!

When they are in motion, the mass's speed will decrease while the plank's will increase. Eventually, they will have the same speed (which means at this instant, the mass is not moving with respect to the plank). No matter what Vo is, they will undergo this event.

I don't understand the question. If it want the mass to stop and reverse its direction, then does that not means ##vo## can be anything? (obviously not 0 since it will make the system not moving at all).

yes

Ah..yes. thank you for reminding me.

This is my answer: $$KE_{total}=KE_{centermass}+KE_{uppermass}+KE_{bottommass}$$ $$KE_{total} = \frac 1 2 (mv^2 + 2m(\vec {v} + \vec {wL})^2) $$ But, the solution manual says that the answer is this: $$KE_{total} = \frac 1 2 (mv^2 + 2m(v^2+w^2L^2)) $$ I think he regard this composite body as...

I have a difficulty when making the energy-conservation-equation for the second step. When making the equation, we need to know the exact position (measured from the sun) of the rocket after it is freed from the Earth gravitation. But, where exactly does the rocket free from Earth...

Ah! So it is not at the equilibrium point! In order to lose contact, plate A should have acceleration g downwards or greater. The only possible location is when the spring is in its natural length or stretched. Am I right?

0. Because it is an equilibrium point. EDIT: Ah. I forgot that B is still in contact with it. So: ##3ma = kx_o -3mg##

g downwards. um... the spring must be at its equilibrium.

In order to solve this problem, we can make use of energy and momentum conservation to solve this problem. But, I'm currently having a difficulty to find out where exactly plate B will lose contact with A. Here is what I'm thinking. First, B will collide inelastically with A, and then they...

Umm... By "internal energies" here, do you refer to the whole system's (slide A,B and mass C)? Pathfinder.

Is this always true if the it is a contact force? Consider this simple scenario. I have an ordinary inclined plane that is free to move and I place a block on it. Soon after I released it, I think the work from the contact force on the mass and on the plane will not cancel out. (This will make...

Now that I think about it... It is pointed out that the slide is much massive than the mass. Then, I think this will not happen if energy is conserved?

Sorry,I have no idea either. That is what the book says. So, I think it is fine to assume that both refer to the same thing ( mass C). I think so. It is not indicated in the book,though. Contact force between slide A and mass C. ##v_2= -m v_1/M##

Firstly, what I'm about to do is to find the velocity of the mass soon after it doesn't touch the slide A anymore. using momentum conservation, I got ##0 = mv' + MV'## with ##v'## and ##V'## are the velocity for mass C and slide A respectively immediately after they lost interaction. Now, I...

Yes

AH! Now I got it. Thanks!

This will be neutralized by torque produced by the pipe's weight. Hence, in the torque equation, we only use friction and normal force.