This is a classic conservation of energy problem. We know that potential energy and kinetic energy both have to conserve. So we use the following equation:

\(\displaystyle \frac{1}{2}mv_o^2+mgh_1 = \frac{1}{2}mv_f^2 + mgh_2\)

What this equations says is, the sum of kinetic and potential energy is same at varying heights and velocities.

We can simplify this equation by cancelling out all the m terms.

\(\displaystyle \frac{1}{2}v_o^2+gh_1 = \frac{1}{2}v_f^2 + gh_2\)

We know all the terms except for \(\displaystyle v_f\), which is the final speed we are trying to solve for \(\displaystyle h_1\), which is \(\displaystyle 24\:m\), \(\displaystyle h_2\) is \(\displaystyle 10\:m\) and \(\displaystyle v_o\) is \(\displaystyle 15\:m/s\).

If we plug in all the numbers and solve for \(\displaystyle v_f\), we get \(\displaystyle 22.3\:m/s\).

Begin by using the following equation relating the initial and final kinetic energy and the work done on the object:

\(\displaystyle KE_i + W = KE_f\)

Then, plug in the given variables and solve for the final speed.

\(\displaystyle \frac{1}{2}mv_i^2 + Fd = \frac{1}{2}mv_f^2\)

\(\displaystyle \frac{1}{2}(25kg)(25\frac{m}{s})^2 + (250N)(13.75m) = \frac{1}{2}(25kg)v_f^2\)

Simplify terms.

\(\displaystyle 7812.5 + 3437.5 = \frac{25}{2}v_f^2\)

Isolate the final velocity and solve.

\(\displaystyle v_f = \sqrt{11250*\frac{2}{25}}\)

\(\displaystyle v_f = \sqrt{900}\)

\(\displaystyle v_f = 30\frac{m}{s}\)

You can use the motion equation and find the maximum, but it may be faster to use energy equations. Set the initial kinetic energy equal to the gravitational potential energy at the maximum height and solve for the height.

\(\displaystyle \frac{1}{2}mv^2 = mgh\)

Mass cancels.

\(\displaystyle \frac{1}{2}v^2 = gh\)

Isolate the height and solve.

\(\displaystyle h = \frac{v^2}{2g} = \frac{(32.65\frac{m}{s})^2}{2(9.8\frac{m}{s^2})} = 54.3889m\)

Round to \(\displaystyle 54m\).

Remember that gravitational potential energy is not affected by the path downward (or upward)—whether it is straight, curved, or winding—only by how big the drop is. Once that is taken into account, you can simply set the initial gravitational potential energy and final kinetic energy equal to each other as if the coaster were falling straight down and solve for the final velocity.

\(\displaystyle mgh = \frac{1}{2}mv^2\)

The mass cancels.

\(\displaystyle gh = \frac{1}{2}v^2\)

Isolate the velocity and solve.

\(\displaystyle v = \sqrt{2gh}\)

\(\displaystyle v = \sqrt{2(9.8\frac{m}{s^2})(45m)}\)

\(\displaystyle v = 29.7\frac{m}{s}\approx30\frac{m}{s}\)

First, find the gravitational potential energy of the drop. Then, set it equal to the kinetic energy at the end of the drop and solve for the velocity.

\(\displaystyle mgh = \frac{1}{2}mv^2\)

The mass cancels.

\(\displaystyle gh = \frac{1}{2}v^2\)

Isolate the velocity and solve.

\(\displaystyle v = \sqrt{2gh}\)

\(\displaystyle v = \sqrt{2(9.8\frac{m}{s^2})(40m)}\)

\(\displaystyle v = 28\frac{m}{s}\)

This gives you the last term you need to solve for the centripetal force.

\(\displaystyle F_c = \frac{mv^2}{r}\)

\(\displaystyle F_c = \frac{(1500kg) (28\frac{m}{s})^2}{200m} = 5880N\)

Set the elastic potential and kinetic energy equations equal to each other:

\(\displaystyle \frac{1}{2}kx^2 = \frac{1}{2}mv^2\)

You are given the fact that in this case, \(\displaystyle x^2 = v^2\). This allows you to simplify the equality.

\(\displaystyle \frac{1}{2}kx^2 = \frac{1}{2}mx^2\)

\(\displaystyle \frac{1}{2}k = \frac{1}{2}m\)

\(\displaystyle k = m\)

This shows us that there is a 1:1 ratio between the spring constant of the elastic and the mass of the object.

We know the maximum height of the pumpkin, which tells us the maximum energy of the launch. Calculate the final gravitational potential energy.

\(\displaystyle PE_g = mgh = (10kg ) (9.8\frac{m}{s^2}) (100m) = 9800J\)

Now set this value equal to a kinetic energy equation that uses the vertical component of the velocity at launch.

\(\displaystyle 9800 = \frac{1}{2}mv_y^2\)

\(\displaystyle v_y = \sqrt{\frac{9800J * 2}{10kg}}\)

\(\displaystyle v_y = 44.27\frac{m}{s}\)

Use the vertical velocity component to determine the launch angle.

\(\displaystyle \theta = \sin^{-1}(\frac{v_y}{v})\)

\(\displaystyle \theta= \sin^{-1}(\frac{44.27\frac{m}{s}}{80\frac{m}{s}}) = 33.60^\circ\approx34^\circ\)

In a perfectly inelastic collision, the two colliding objects stick together; the two colliding objects deform, but mass is still conserved. Momentum is conserved during collisions of any sort, including inelastic collisions.

Kinetic energy is reduced during an inelastic collision, and is only conserved in elastic collisions. During inelastic collisions, some kinetic energy is lost to the environment in the form of heat or sound.

The problem does not give any information regarding position, and thus we cannot comment on any changes or lack of changes in potential energy.

The primary difference between elastic and inelastic collisions is the conservation of kinetic energy. Kinetic energy is conserved in elastic collisions, but is not conserved in inelastic collisions. Momentum is always conserved, regardless of collision type. Mass is conserved regardless of collision type as well, but the mass may be deformed by an inelastic collision, resulting in the two original masses being stuck together.

There is no description in the problem reagrding position, so we cannot comment on potential energy.

The collision is assumed to be elastic, so both momentum and kinetic energy are conserved. Use the law of conservation of momentum:

\(\displaystyle \sum p_i=\sum p_f\)

Momentum is the product of velocity and mass:

\(\displaystyle p=mv\)

We can expand the summation for the initial and final conditions:

\(\displaystyle m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}\)

Use the given values to fill in the equation and solve for \(\displaystyle v_{1f}\):

\(\displaystyle (200kg)(30\frac{m}{s})+(150kg)(0\frac{m}{s})=(200kg)v_{1f}+(150kg)(10\frac{m}{s})\)

\(\displaystyle 6000\frac{m\cdot kg}{s}=(200kg)v_{1f}+1500\frac{kg\cdot m}{s}\)

\(\displaystyle v_{1f}=\frac{6000\frac{m\cdot kg}{s}-1500\frac{kg\cdot m}{s}}{200kg}=22.5\frac{m}{s}\)

Since the final velocity is positive, we know that the car is still traveling toward the west.