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8

There's a simple way to look at this that doesn't involve any maths. Suppose the two cars are parked and are stationary, and you accelerate past them in your car. If you are accelerating forwards then from your perspective it looks as if the two cars are accelerating backwards (at the same rate). But the cars are at rest, so the distance between them can't ...


4

It makes no sense for a point mass to have 2 accelerations. What you might have done is find accelerations due to 2 forces separately. You can add them as when $m= \text{constant}$, $\vec{F}=\vec{F_1}+\vec{F_2}=m(\vec{a_1}+\vec{a_2})$ When using vectors symbol, its automatically takes care of their directions.


3

1) Yes indeed, the absence of a $\Delta$ in the second expression is just a typo. 2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity $ 4v $ changes, we really only need to know how the quantity $v$ changes. Suppose $v$ changes from $v_1$ to $v_2$. ...


3

As Sachin says, there is no limit to acceleration. In fact you can show this by considering an observer hovering at a fixed distance from a black hole. As described in this question, the acceleration required to maintain a fixed distance from a black hole is given by: $$ a = \frac{GM}{r^2}\frac{1}{\sqrt{1-\frac{r_s}{r}}} $$ where $r_s$ is the radius of the ...


2

This is due to the superposition principle: when several forces act upon a body, the net force is the sum of the individual forces: $$\vec F_{net} = \sum \vec F_i $$ However, this is only true when the relation between the force and the acceleration is linear. Let's take the gravitational force as an example: say you have three bodies and you have already ...


2

There is no law that says the sum of forces on a given object must be $0$, that is simply the condition for mechanical equilibrium. If an object has constant $0$ velocity (or, more generally, any constant velocity), then its acceleration ($\frac{dv}{dt}$) is $0$ and, by Newton's second law as you have it, the net force acting on it is $0$. However, if all ...


2

Lets do a Free Body Diagram of the lift (NOTE: Always do A FBD first). What are the forces acting on the lift? $$\sum \vec{F} = \begin{pmatrix} -T \sin\psi \\ T \cos\psi - W \end{pmatrix} $$ What is the acceleration on the lift? $$\vec{a} = \begin{pmatrix} -a \cos \theta, a \sin \theta \end{pmatrix} $$ Combine them with $\sum \vec{F} = m \vec{a}$ and ...


2

The vertically moving object is an Atwood machine and the two masses have their own accelerations that are in different directions. The acceleration of $m_2$ and $m_3$ (separate from the total system) is given by $$ a=\frac{m_3-m_2}{m_2+m_3}g\tag{1} $$ Mass $m_2$ is accelerating upwards, hence the acceleration in your case of $a_0-a$; likewise mass $m_3$ is ...


2

A force is applied to a box on a table(lets ignore friction), and the box moves with some constant velocity. It's impossible. Or, don't ignore friction. When an object moves with constant velocity, the total net force on the object is always zero. If you have applied force, there's another force (or, many forces) like friction to counterbalance it. ...


2

Maybe you are pouring sand on your box.$$F=\frac{dp}{dt}=v\frac{dm}{dt}+m\frac{dv}{dt}$$ $$\text{As, } v=0 ms^{-1}$$$$F=v\frac{dm}{dt}$$ Second possibility : If your box is spherical, By Stokes' Law $$F_{viscous}=6\pi\eta rv$$ where $\eta$ is coefficient of viscosity. Hence, your ball attains terminal velocity. $$F=6\pi\eta rv$$ $$v=\frac{F}{6\pi\eta ...


2

By Newton's second law of motion, if there is a nonzero net force there is an acceleration. If there is no acceleration then the net force is zero. In the situation you describe, where the box has no acceleration, there must be another force balancing $F_{app}$ otherwise there will be an acceleration.


1

The problem in yours is that you are taking the net force acting downward to be $(m_2+m_3)g$ is incorrect and that led you to take the total mass to be $m_1+m_2+m_3$ which is again incorrect because $m_2\neq m_3$. If $m_2=m_3$ then the center of mass of $m_2$ and $m_3$ will lie on the straight vertical line through the center of the pulley B and the force ...


1

While the other answer are all completely correct, I just want to write a more simplified answer. It's much the same as distances. I you walk 1 meter North and 1 meter East, you can add the two distance vectors and get $\sqrt2$m North-East: $$\vec{d}_1=1m[N]=(1,0),~~\vec d_2=1m[E]=(0,1)$$ $$\vec d=\vec d_1+\vec d_2=(1,1)=1m[N]+1m[E]=\sqrt2m[NE]$$ Adding ...


1

Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be ...


1

1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere ...


1

Well that's a result using differentiation and derivation. Have you studied calculus? If not, there is a simple way to look at it. $$\frac{\Delta p}{\Delta t} = \frac{\Delta (mv)}{\Delta t}$$ (Yes there should be a $\Delta t$ in the denominator, too.) Now, what does $\Delta(mv)$ mean? It represents the change in the quantity $mv$. For current situations, ...


1

You are right, there is a $\Delta$ missing in front of the $t$. $\Delta v = v_2 - v_1$. If the mass is not changing, then $\Delta (mv) = mv_2 - mv_1 = m(v_2 - v_1) = m\Delta v$. Hope that helps. The equation that includes $\frac{\Delta m}{\Delta t}$ is not Newton's second law. The second law is valid only for systems of constant mass. An equation like ...


1

Yes. It should be: $$\frac{dp}{dt}=\frac{d(mv)}{dt}$$ I'm using $d$ instead of $\Delta$ because I am thinking about the limit where the changes in $p$ and $t$ are very small. Then these are called infinitessimal changes, and denoted by a $d$. Usually, when one considers simple problems in Newtonian mechanics, what one does is study a given object with a ...


1

No, it is not correct. Let $a$ be acceleration; $F$ force; $m$ mass; $v_0$ initial velocity; $v_f$ final velocity; $P$ power required; $x$ distance travelled and $t$ time taken. Hence, $$P=\frac{Fs}{t}=\frac{ma\ x}{t}$$ Then $a=\frac{v_f^2-v_0^2}{2x}=\frac{5^2-0^2}{2*10}=1.25\text{ m s}^{-2}$ and $t=\frac{2x}{v_0+v_f}=\frac{2*10}{0+5}=4\text{ s}$ Hence, ...



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