Your comments

If the ball is being released from a different height than the net stands, then you can't use the formula for the maximum range. Use your ordinary kinematic equations to find the initial speed.

You have to be careful how you use conservation of energy because this problem has a collision. Energy is not always conserved in collisions! On the other hand, momentum is always conserved.


You can use energy to find the velocity of the skier at the bottom of the slope just before the collision, then use conservation of momentum to find the velocity of the skier + fiance combination right after the collision. Then you can try using energy again to find the maximum height they can get up the hill.

This is just a kinematic problem, but it looks more complicated because there are two spaceships. Really, we have a kinematic problem in terms of the distance between the spaceships and the relative velocity between them.


The first ship is traveling at 50 km/s, and the second at 20 km/s, so the relative velocity between them is 30 km/s. There is a separation of 100 km between them, so without decceleration the gap will be closed in


100 / 30 ~ 3.33 seconds.


If the first ship deccelerates, then we want the relative velocities of the ships to be zero after 100 km.


We can formulate the kinematic problem as follows:


distance= 100

v_initial = 30

v_final = 0


Is there a kinematic equation that lets you solve for acceleration from these quantities?

The reason there is a definite mass of the second ball is that the collision is elastic, meaning that energy is also conserved in addition to momentum. Now we have two equations,


p_initial = p_final

KE_initial = KE_final


and two unknowns (the masses), so we will be able to solve for the second ball's mass.

Be careful, because this problem has multiple steps. First, you have a kinematic problem to determine the bullet's speed right before it collides with the block. Then you have the collision itself in which momentum must be conserved. Then you have another kinematic problem where the block and bullet combination rises into the air and reaches its peak.


Make sure you treat each part of the problem separately and check your results before moving on to make sure they make sense. The bullet should lose a little bit of speed before it collides with the block. The momentum of the bullet before the collision is equal to the momentum of the block and bullet combination, which now has a combined mass that is much greater. Then the velocity of the block and bullet combination after the collision acts as your initial velocity for the final kinematic stage.

The formula for range of a projectile is


R = v^2 / g * sin(2θ), where v is the initial velocity


So 45 degrees does get you the maximum horizontal range, but I don't think that's what the question is asking. It says to set the range equal to the maximum height, which means you might have to solve for the optimum angle.


Try using energy to help solve this problem. If the projectile is fired from the ground, the initial energy is just the kinetic energy: 1200 J. When the bullet reaches its maximum height, its energy consists of gravitational potential energy and kinetic energy (horizontal motion only). Try writing out and setting these equal:


1200 J = mgh + 1/2 m (v_h)^2


where h is the maximum height, and v_h is the velocity at the maximum height. We can get somewhere with this equation by substituting in h = R, and noting that v_h is just the x-component of the initial velocity because it doesn't change, v_h = v cos(θ)


You can solve for the initial velocity v using the initial kinetic energy provided. Then, you should have an equation in terms of the angle. You might have to use trig identities to solve for the angle, but once you do you can then solve for the range using the original formula.

Magnetic fields are created from moving charges, or currents. In atoms, electrons typically "orbit" the nucleus and thus produce a current and magnetic field. We call this a magnetic dipole because it has North and South poles. Dipoles like to attract and align with each other, and this is why two bar magnets, for example, will try to align their poles North to South.


Magnetization refers to the sum of all the dipoles (per volume). A strong magnet has a high magnetization because it has many dipoles all aligned.

The formula you provided is a differential, meaning you're finding the force dF on a small piece of wire dl. This bit of wire is infinitesimally small, so B and I are essentially constant.


If B is not constant then you will need to take an integral over the path of the wire. In general this is not what we do to solve problems, though, so usually you use F = I L x B for a straight piece of wire L and constant B

First, think about two objects moving in a straight line at the same speed. Which has more linear momentum? The formula is p = mv, so clearly the object with more mass has more momentum.


Now, think about two spinning objects. The formula for angular momentum is L = I ω, where I is the moment of inertia and ω is the angular velocity. From the equation we can see that, of the two objects spinning at the same rate, the one with the higher moment of inertia has more angular momentum.


Moment of inertia is a measure of the distribution of the mass. If more of the mass is farther away from the spin axis, the moment of inertia is higher. So for the case of a solid sphere vs a hollow sphere with the same masses, a hollow sphere has its mass all distributed at the radius whereas the solid sphere is more uniformly distributed. The hollow sphere then has a higher moment of inertia and thus more angular momentum.

A capacitor can be as simple as two metal plates. When a potential difference V is applied to a capacitor, current will flow, meaning charge is forced to pile onto the plates: positive on one end and negative on the other.


We know, however, that like charges repel each other, so the charges will resist piling up. Eventually, the plates will reach the maximum amount of charge that can be stored on the plates for a given potential V. It is the repulsion of like charges that prevents the charging from being instantaneous.