52

Your reasoning would be correct if the swimmer in the river was trying to reach the point on the bank opposite where they started. To do this they have to swim in a direction angled upstream, so relative to the water they have to swim a longer distance than the width of the river. But to cross the river in the minimum time the swimmer should swim in a ...


19

Projectile motion is motion under constant acceleration. Projectiles move on a parabolic path where both the magnitude and the direction of the velocity are continuously changing but the magnitude and the direction of the acceleration are not changing. The textbook is correct.


8

If we find the force for a decelerating object (let's say a car on a road), we'll see that the force is negative as in it’s in the opposite direction of the motion of the object. Not necessarily. The sign of the force just depends on its direction; it has nothing to do with the direction of motion of the object. If the object is moving in the positive ...


8

This is possible. Let us suppose a body moves in the positive $x$ direction with a constant velocity $v_x$. We provide a constant acceleration in the $y$ direction equal to $a$. After a time $t$, the velocity of the body will be $v = v_x \hat{x} +at\hat{y}$. The direction changed by $\theta = \tan^{-1}\frac{at}{v_x}$. This happens in many cases, and one ...


7

Let's leave aside the evident mistake in the formulae ($L/g \rightarrow g/L$) so that the formula should be read as $$ \Delta T = T-T_0=\frac{2 \pi}{\sqrt{g}}\left(\sqrt{L+\Delta L}- \sqrt{L} \right), $$ What the textbook is saying is that the formula gives little insight as to how $\Delta T$ depends on $\Delta L$. I wouldn't consider equivalent little ...


6

As gandalf61 has explained, one way to view this problem is the current flow could be considered orthogonal to the intended direction of travel. With such a view, the minimum crossing time solution is to swim directly perpendicular to the current, while allowing the current to sweep you downriver. But I will make what is sure to be a contentious claim: this ...


6

You can consider the addition of mass to be an inelastic collision where kinetic energy is not conserved by the momentum is conserved. The linear momentum in the horizontal direction of motion should remain the same before and after the collision. $$mu = (m+M)v$$ where $m$ is the mass of the trolley, $M$ is the added mass, u is the initial velocity of the ...


5

I think both statements are correct. In general, given some random physical system with no particularly special properties, we should say that the energy (including the potential energy) is a property of the system. The potential energy is generally defined to be the energy that would be needed to construct the system, starting from pieces that are non ...


4

A force (F) is a vector quantity. A vector has a magnitude (which cannot be negative) and a direction. The negative of (F) is a different vector with the same magnitude but opposite in direction. When working with vectors, it is usually best to work with their components. In a chosen coordinate system, a vector may have one or more negative components ...


4

then the force from the person (their weight) would act upon the egg. That's not true. You can push on the egg, but that doesn't mean that you're actually applying that much force. Push a wall, you can probably generate 100N or more against the wall. Now push a feather. You will not be able to apply that much force. However, if the egg can only handle a ...


4

An object can have a changing direction of velocity, and this is still called an acceleration, but the acceleration itself need not change (constant acceleration, meaning its magnitude and direction stay the same). Consider an object moving as a projectile on earth's surface $$\vec x(t) = \vec v_0 t+\frac{1}{2}\vec gt^2$$ $$\vec v=v_x\hat i+gt\hat j$$ At any ...


3

There's a few pieces to this. The first is that it will be easier to understand forces if you don't think of them as "negative." Forces are vectors, which have a (positive) magnitude and a direction. Where you can get a "negative" thing regarding vectors is when you write them down in component form. When using components you pick ...


3

You're kind of overthinking this. Newton's 3rd law does not say there have to be two forces for every interaction, it's stating a fact about the nature of what a force is. It is essentially saying there is no such thing as a "one sided" force -----> All forces are like this . <-----> In your example, what happens is the person ...


3

Momentum is always conserved. Perhaps you are thinking of a system where the mass is brought up to speed with the trolley, then gently attached to it? In that case, then the trolley will, indeed, have more momentum than before the addition -- precisely as much as the mass had by itself! If the trolley crashes into the mass, then you're talking about the &...


3

Your system is the mass and trolley, so that before the mass is added, the total momentum is that of the trolley only since the mass is presumably not moving. Now when the mass is added (gently so that there is no downward velocity component), the trolley will move with a different velocity, but the total momentum of the trolley and mass system will be ...


3

If you are working in a rotating reference frame you need to include a centrifugal force term. If you are working in an inertial reference frame then you do not include it. This is true of any pseudo-force. They do not arise from the motion of the object; they arise from the acceleration of the reference frame. To address another point, the spring will not ...


3

One way this can be done is by parametric excitation: if you raise and lower the pivot point slightly at twice the natural frequency of the pendulum, such that it is highest when the bob is at the bottom and lowest when the bob is at either end of its swing, it puts energy into the oscillation. Since the imposed motion is purely vertical, it doesn’t alter ...


2

Let's try to understand it using an example: Initially if a body had a certain velocity in the positive x direction and the acceleration in the negative x direction then as the time progresses the velocity of the body would keep decreasing. Eventually it would become zero followed by a velocity in the negative x direction. So, the direction of its velocity ...


2

You can calculate moments of inertia about any axis you want. But if the system isn't actually rotating around that axis, then you can't use it to (directly) calculate the rotation speed. Instead of all the parts moving with speed proportional to $r$ as the distance from your axis, the parts are moving with various speeds as it spins around the center of ...


2

This part is incorrect: $$E_i = F_i d = (2 m_i \frac{d}{t^2}) d = 2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$ Specifically $$2 m_i \frac{d^2}{t^2} = 2 m_i v_i^2$$ is wrong because by your own omission $v_i \neq \frac{d}{t}$


2

The conceptual problem is balancing forces, when it requires balancing momenta. Of course: $$ F = \frac{dp}{dt} $$ The gravitational force is indeed constant, so after falling for a time $t$: $$ p = \int_0^t F(t')dt'=F\int_0^t dt'=Ft$$ He needs to apply a force to the box that gives him an impulse (the product for force times time) of $-p$.


2

To understand the conservation of momentum here, it is necessary to think about the forces involved, because the principle of conservation of momentum has a dependent clause: If the net external force on a system is zero, then the momentum is conserved. Because momenta and forces are vector quantities, this actually holds true for each component of an ...


2

I think you have misunderstood the question. What you have calculated by $F=mg$ (with $g=9.81\text{ m/s}^2$) is just the vertical weight force of the vehicle. But the question is not about friction or vertical force. Instead the question describes the horizontal movement of a vehicle. You can set up the kinematic equations for position $x(t)$ and velocity $...


2

Best to draw the graph. The distance travelled (hence position) is equal to the area under the curve.


2

Another method often used by museums is by having iron collars wrapped around the cable close to the pivot point at the top. There will be a torus-shaped electromagnet built into the ceiling, and the cable with the iron collars swings back and forth inside this torus. Just before the pendulum reaches its maximum height, the electromagnet is given a short-...


2

You're right, Frictional $F$ does no work on the runner in the ground frame. Running is a more complex motion. During the thrust phase of your step, the muscles in your leg are doing work on your moving torso. ($F$ holds your foot still so this work can be done). Then during the recovery phase, your leg muscles do work on your foot to lift and accelerate ...


2

To complement other answers without directly invoking Newton’s law, the asymmetry comes from energy. The asymmetry between car and Earth comes from the fact that car has to consume and spend energy (do work) in order to change its position. —— Actually we can even forget about energy and go further by saying that car transfers its momentum to earth. Imagine ...


2

Your question and your title are different. Understanding that difference may help! We'll start with the question, because it has an easy thought experiment available. Start with the car moving very slow. Maybe it ran out of gas and someone is pushing it. In the first case, it is pushed off of the cliff. In the second, someone has set up a smooth ramp ...


2

Option D is incorrect. If the man's foot is stationary with respect to the plank then the friction between the man and the plank is static friction. If the man's foot is not stationary with respect to the plank (i.e. it is sliding along the plank) then the friction between the man and the plank is kinetic friction. In either case, the type of friction ...


2

Although it might look counterintuitive at first, block $A$ will indeed become very fast. To see this in reality, you can take a string, maybe 0.5 m long, hold both ends in your hands so that the string is loose and then pull them apart. Even if you do not move your hands extremely fast, the string will start to vibrate when it becomes taut, fast enough to ...


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