All High School Physics Resources
Example Questions
Example Question #1 : Conservation Of Momentum
A ball moving at strikes a ball at rest. After the collision the ball is moving with a velocity of . What is the velocity of the second ball?
This is an example of an elastic collision. We start with two masses and end with two masses with no loss of energy.
We can use the law of conservation of momentum to equate the initial and final terms.
Plug in the given values and solve for .
Example Question #1 : Conservation Of Momentum
A object moves to the right at . It collides head on with a object moving to the left at . Which statement is correct?
The total momentum before the collision is and after the collision is
The total momentum before and after the collision is zero.
The total momentum both before and after the collision is
None of the above is true
The total momentum before and after the collision is zero.
The total momentum before the collision is equal to the momentum of each object added together.
Remember that moving to the left means that the object has a negative velocity
Total momentum =
According to the law of conservation of momentum, the total momentum at the end must equal the total momentum at the beginning. Since the momentum at the beginning was , the momentum at the end would also be .
Example Question #1 : Elastic And Inelastic Collisions
Two bumper cars at an amusement park collide elastically as one approaches the other directly from the rear. The car in front (CarA) has a mass of and the car behind it (CarB) has a mass of . The car in front was traveling at while the car behind hit him with a velocity of . What are their final velocities after the collision?
Knowns
Unknowns
For elastic collisions we know that the initial and final velocities are related by the equation
We also know that the momentum is conserved meaning that
Since we have two missing variables and two equations, we can now solve for one of the variables using a system of equations
Let’s get the final velocity of car A by itself from the first equation
We can now substitute this equation into our momentum equation.
In our new equation we only have one missing variable, so lets substitute in values and solve.
We can now plug this value back into our equation for
Example Question #2 : Elastic And Inelastic Collisions
You are a witness in a court case involving a car accident. The accident involved car A of mass which crashed into stationary car B of mass . The driver of car A applied his brakes before he crashed into car B. After the collision, car A slid while car B slid 3. The coefficient of kinetic friction between the locked wheels and the road was measured to be . What was the velocity of car A before hitting the brakes?
Knowns
Unknowns
To solve the problem we must consider three different situations. The first is when the cars are skidding across the ground. While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.
The second situation is the collision itself when the first car has an initial speed and the second car is stopped. After the collision both cars are traveling with the different speeds as they skid across the ground at different distances.
The third situation is when the first car initially hits the brakes before the collision. While he is skidding to slow down, the force of friction is resisting the motion and therefore doing work to slow the car down some.
To solve this problem we must work from the end of the collision and work backwards to find out what happened before.
To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.
The work-kinetic energy theorem states that the work done is equal to the change in kinetic energy of the object.
Work is directly related to the force times the displacement of the object.
In this case, the force that is doing the work is friction.
Since the cars are on a level surface the normal force is equal to the force of gravity.
The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.
When we put all these equations together we get
We also know that kinetic energy is related to the mass and velocity squared.
Therefore our final equation should look like
Notice that the mass falls out of the equation since it is in every term. Also note that the final velocity is 0m/s so this will cancel out as well.
Since both of these terms have a negative we can cancel this out as well.
We can now substitute our variables to determine the velocity of each of the cars just after the crash.
This is the velocity of each of the cars after the collision. We must now consider our second situation of the collision itself. During this collision momentum must be conserved. The law of conservation of momentum states
We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found) and the initial value of the stopped car.
Now we can solve for our missing variable.
Now we must consider our third situation which is when car A is braking. We can return to our work-kinetic energy theorem equation again as the force of friction is what is slowing the object down.
This time however, the final velocity is not 0m/s as the car A is still moving when he crashes into car B. The final velocity is the velocity that we determined the car was moving with before the collision in the second part of the problem. The masses do still cancel out in this case.
Example Question #3 : Elastic And Inelastic Collisions
Hockey puck A () travels with a velocity of to the right when it collides with hockey puck B (), which was originally at rest. After the collision, puck A is stationary. Assume no external forces are in play and that the momentum of the pucks are conserved. What is the final velocity of puck B after the collision?
Elastic collisions occur when two objects collide and kinetic energy isn't lost. The objects rebound from each other and kinetic energy and momentum are conserved. Inelastic collisions are said to occur when the two objects remain together after the collision so we are dealing with an elastic collision.
Above, the subscripts 1 and 2 denote puck A and B respectively, and the initial momentum of puck B is zero, so that term is not included in the equation above.
Plug in initial and final velocities and mass:
Example Question #1 : Elastic And Inelastic Collisions
A car with mass and initial velocity strikes a car of mass , which is at rest. If the two cars stick together after the collision, what is the final velocity?
We know that the cars stick together after the collision, which means that the final velocity will be the same for both of them. Using the formula for conservation of momentum, we can start to set up an equation to solve this problem.
First, we will write the initial momentum.
We know that the second car starts at rest, so this equation can be simplified.
Now we will write out the final momentum. Keep in mind that both cars will have the same velocity!
Set these equations equal to each other and solve to isolate the final velocity.
This is our answer, in terms of the given variables.
Example Question #5 : Elastic And Inelastic Collisions
A sports car collides into the rear end of SUV stopped at a red light. The bumper lock, the brakes lock and the two cars skid forward before coming to a stop. The police officer, knowing the coefficient of kinetic friction between the tires and the road is about . He calculates the speed of the sports car at impact. What was that speed?
Knowns
Unknowns
To solve the problem we must consider two different situations. The first is when the cars are skidding across the ground. While they are skidding the force of friction is what is resisting their motion and therefore doing work on the cars to slow them down to a stop.
The second situation is the collision itself when the first car has an initial speed and the second car is stopped. After the collision both cars are traveling with the same speed as their bumpers have been locked together.
To begin, we need to find the speed that the cars are skidding across the ground after the bumpers have been interlocked.
The work-kinetic energy theorem states that the work done is equal to the change in kinetic energy of the object.
Work is directly related to the force times the displacement of the object.
In this case, the force that is doing the work is friction.
Since the cars are on a level surface the normal force is equal to the force of gravity.
The force of gravity is directly related to the mass and the acceleration due to gravity acting on an object.
When we put all these equations together we get
We also know that kinetic energy is related to the mass and velocity squared.
Therefore our final equation should look like
Notice that the mass falls out of the equation since it is in every term. Also note that the final velocity is 0m/s so this will cancel out as well.
Since both of these terms have a negative we can cancel this out as well.
We can now substitute our variables to determine the velocity of both cars just after the crash.
This is the velocity of the cars after the collision. We must now consider our second situation of the collision itself. During this collision momentum must be conserved. The law of conservation of momentum states
We can substitute our values for the masses of the cars, the final velocity of the cars after the collision (which we just found) and the initial value of the stopped car.
Now we can solve for our missing variable.
Example Question #881 : High School Physics
A car traveling at collides with another car that is at rest. The two bumpers lock and the cars move forward together. What is their final velocity?
This is an example of an inelastic collision, as the two cars stick together after colliding. We can assume momentum is conserved.
To make the equation easier, let's call the first car "1" and the second car "2."
Using conservation of momentum and the equation for momentum, p=mv, we can set up the following equation.
Since the cars stick together, they will have the same final velocity. We know the second car starts at rest, and the velocity of the first car is given. Plug in these values and solve for the final velocity.
Example Question #6 : Elastic And Inelastic Collisions
Which of these would NOT be an example of an inelastic collision?
A man is clapping his hands such that they move with equal, but opposite velocities
A match scrapes a matchbook and bursts into flame
Two hydrogen atoms fuse together to form a helium atom and gamma radiation
Two cars crash into each other, and stop with a loud bang
Neutrons fuse with hydrogen atoms in a nuclear reactor core such that kinetic energy is conserved
Neutrons fuse with hydrogen atoms in a nuclear reactor core such that kinetic energy is conserved
The difference between an elastic and an inelastic collision is the loss or conservation of kinetic energy. In an inelastic collision kinetic energy is not conserved, and will change forms into sound, heat, radiation, or some other form. In an elastic collision kinetic energy is conserved and does not change forms.
Remember, total energy and total momentum are conserved regardless of the type of collision; however, while energy cannot be created nor destroyed, it can change forms.
In the answer options, only one choice preserves the total kinetic energy. The resulting bang from the car crash, the flame from the match, the sound of hands clapping, and the gamma radiation during hydrogen fusion are all examples of the conversion of kinetic energy to other forms, making each of these an inelastic collision. Only the neutron fusion described maintains the conservation of kinetic energy, making this an elastic collision.
Example Question #882 : High School Physics
A spring has a spring constant of . If a force of is used to stretch out the spring, what is the total displacement of the spring?
For this problem, use Hooke's law:
In this formula, is the spring constant, is the compression of the spring, and is the necessary force. We are given the spring constant and the force, allowing us to solve for the displacement.
Plug in our given values and solve.
Note that both the force and the displacement are positive because the stretching force will pull in the positive direction. If the spring were compressed, the change in distance would have been negative.