Momentum is always conserved in physical collisions. Momentum is represented by the equation:

\(\displaystyle p=mv\)

When the cars collide the momentum of each car contributes to a total momentum of a new object that is a combined mass of the original two cars. The equation for this would appear as follows where car 1 is object one, car 2 is 2, and the object after the collision is object 3:

\(\displaystyle m_{3}v_{3}=m_{1}v_{1}+m_{2}v_{2}\)

Because the mass of the first car is 1250 kg and the mass of the 2nd car is 1400kg, their combined mass for the resulting object is 2650 kg. By plugging in values, the velocity of this combined object is found to be:

\(\displaystyle 2650v_{3}=1250(60)+1400(30)\)

\(\displaystyle v_3\approx 44\ \frac{m}{s}\) 

Due to conservation of momentum, all of the momentum that the ball has must be conserved after the collision. Because the cue ball completely stops moving after the collision it has no momentum meaning that all of the momentum is transferred to the 8 ball. Momentum is represented by this equation:

\(\displaystyle p=mv\)

Since the momentum must be equal before and after the collision, this equation can be formulated for this problem:

\(\displaystyle m_{1}v_{1}=m_{2}v_{2}\)

Where the cue ball is object 1 and the 8 ball is object 2. By plugging in the given values it is possible to find the velocity of the 8 ball after the collision. 

\(\displaystyle 0.142*26=0.167v_{2}\)

Solving for \(\displaystyle v_{2}\) gives a velocity for the 8 ball of \(\displaystyle 22\ \frac{m}{s}\)

In order to start this problem, a free body diagram illustrating all the forces acting on the object must be drawn. Because there are only 2 forces acting on the object the free body diagram would appear as follows (not to scale):

T1 stands for the tension in the rope created by body builder 1, T2 stands for the tension created by body builder 2. The total net force acting on the object, in the direction of body builder 1 is equal to

\(\displaystyle 1115N-920N=195N\)

Newton's second law relates force, mass, and acceleration with the following equation:

\(\displaystyle F=ma\)

By plugging in the given values for force and mass, acceleration is found to be:

\(\displaystyle 195=12a\)

\(\displaystyle a=16.25\ \frac{m}{s^{2}}\)

The force-momentum relationship is:

\(\displaystyle \vec{F} \Delta t=\Delta \vec{p}\)

Since both force and change in time are located on the left side of the equation, they are inversely proportional. Therefore, if the force doubles, the inverse must be true about the time. This would halve the time necessary for the cars to bump. 

This is a conservation of momentum problem:

\(\displaystyle m_sv_s+m_hv_h=m_fv_f,\) where the subscripts refer to \(\displaystyle Sally(s),\) \(\displaystyle hoverboard(h)\), and \(\displaystyle final(f).\)

Now, let's plug in the relevant information. Remember, the hoverboard is initially stationary, so its velocity is zero. Also, for the final mass, we need to add the two masses, since now Sally is on top of the hoverboard.

\(\displaystyle m_sv_s+m_hv_h=m_fv_f=20*5+5*0=25v_f.\)

Now, all we need to do is solve for the final velocity:

\(\displaystyle v_f=\frac{20*5}{25}=\frac{100}{25}=4\frac{m}{s}\)

For this question, we're told that a runner begins at rest and travels a certain distance with a constant acceleration. We're asked to determine how much time it takes for this to happen.

Right off the bat, it's important to realize that we can use the kinematics equations because we are told that the acceleration is constant. Now, we need to use an equation that relates distance, time, and acceleration.

\(\displaystyle \Delta x=v_{i}t+\frac{1}{2}at^{2}\)

Since we know that the person is starting from rest, their initial velocity must be zero. Thus, we can simplify the above expression.

\(\displaystyle \Delta x=\frac{1}{2}at^{2}\)

Now, we can go ahead and rearrange this expression to isolate the variable for time.

\(\displaystyle t=\sqrt{\frac{2\Delta x}{a}}\)

Lastly, we just need to plug in the values given to us in the question stem in order to solve for our answer.

\(\displaystyle t=\sqrt{\frac{2(5\:m)}{2\: \frac{m}{s^{2}}}}\)

\(\displaystyle t=2.24\:s\)

In this question, we're told that two objects of equal mass eventually collide with each other and then travel together. We're asked to find an expression that tells us the final momentum of the system.

First, we can recognize this as an inelastic collision, which means that momentum is conserved in our system of the two objects but kinetic energy is not. Also, because both objects travel together after the collision, there will be a single final velocity.

\(\displaystyle mv_{1}+mv_{2}=(m+m)v_{f}=2mv_{f}\)

Thus, the final momentum of our system of two objects can be expressed as \(\displaystyle 2mv_{f}\).

At a projectiles maximum height its y-direction velocity is always \(\displaystyle 0\frac{m}{s}\). It is moving neither up nor down. Because the ball was thrown straight in the air, there is no x-direction component of its velocity to account for.

Gravity is the only force acting on the ball at its maximum height. Though the ball has zero velocity at this point, gravity is still accelerating the ball at \(\displaystyle -10\frac{m}{s^2}\). Gravity almost always pulls down, which is a negative direction in the traditional coordinate system so our answer must be negative.

We can calculate the distance traveled by multiplying the velocity of the ball by the reaction time of the batter. However, we will need to first need to get the appropriate units:

\(\displaystyle v = 102mph \left ( \frac{5280feet}{1mile}\right )\left (\frac{1meter}{3.28084feet}\right )\left ( \frac{1hour}{3600seconds}\right )\)

\(\displaystyle v = 45.6\frac{m}{s}\)

Now we can simply multiply this by the reaction time of the batter to get the distance that the ball has traveled by the time her reacts:

\(\displaystyle d = v\cdot t = \left ( 45.6\frac{m}{s}\right )(0.20s)\)

\(\displaystyle d = 9.1m\)

Fun fact: the distance from the pitching mound to home plate is about \(\displaystyle 18.4m\), so the ball is about halfway to the plate by the time the batter begins to react.