All High School Physics Resources
Example Questions
Example Question #1 : Understanding Conservation Of Momentum
A car travelling at rear ends another car 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, , 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 #11 : Momentum
A car strikes a car at rest from behind. The bumpers lock and they move forward together. If their new final velocity is equal to , what was the initial speed of the first car?
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, , 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 final velocity is given. Plug in these values and solve for the initial velocity of the first car.
Example Question #1 : Understanding 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 after the collision?
We can use the law of conservation of momentum:
We know the mass of each ball and their initial velocities.
We also know the final velocity of the first ball. This leaves only one variable: the final velocity of the second ball.
Solve to isolate the variable.
Example Question #3 : Understanding Conservation Of Momentum
A ball moving at strikes a second ball at rest. After the collision the ball is moving with a velocity of and the second ball is moving with a velocity of . What is the mass 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 the mass of the second ball.
Example Question #3 : Understanding Conservation Of Momentum
A ball strikes a second ball at rest. After the collision the ball is moving with a velocity of and the second ball is moving with a velocity of . What is the initial velocity of the first 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 the initial velocity of the first ball.
Example Question #1 : Understanding Conservation Of Momentum
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 #1 : Understanding Conservation Of Momentum
Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different?
No difference
They would rebound at a higher speed
They would rebound at a slower speed
No difference
Consider the law of conservation of momentum.
If both balls are identical then we can say that
Therefore we can state the equation as
Since is the common factor we can remove it from the equation completely.
Since mass factors out of the equation; then it does not matter if the balls increase or decrease in their mass.
Example Question #6 : Understanding Conservation Of Momentum
You are lying in bed and want to shut your bedroom door. You have a bouncy ball and a blob of clay, both with the same mass. Which one would be more effective to throw at your door to close it?
The blob of clay
Both the same
The bouncy ball
Neither will work
The bouncy ball
A bouncy ball will have an elastic collision with the door, causing the ball to move backward at the same speed it hit the door.
On the other hand, the blob of clay will have an inelastic collision with the door, causing the blob of clay to move with the same speed as the door.
So let us look at the conservation of momentum for an elastic collision.
Since the door has no initial velocity we can remove it from the beginning of the equation.
Since the ball will rebound with the same amount that it hits the door the velocity at the end is a negative of the velocity at the beginning.
Rearrange for the final velocity of the door.
Now let us examine the law of conservation of momentum for the inelastic collision. Again the door has no initial velocity so we can remove it from the beginning of the equation.
Since the collision is inelastic, the final velocity of both objects will be the same so we can set them equal to each other.
Now solve for the final velocity of the door.
When we compare these two final velocities, it is clear that the elastic collision will create a larger velocity for the door because the top number is twice the value as the inelastic collision, and it is being divided by only the mass of the second object, instead of both objects combined.
Example Question #281 : Motion And Mechanics
A 12kg hammer strikes a nail at a velocity of 7.5m/s and comes to rest in a time interval of 8.0ms What is the average force acting on the nail?
11250N
8750N
15350N
1150N
11250N
Knowns
First, let us find the change in the momentum of an object.
We know that the change in momentum of an object is equal to the impulse of an object. Impulse is equal to the force acting on the object multiplied by the time that force is applied.
We need to convert our time from milliseconds to seconds.
Solve for the force
Example Question #281 : Motion And Mechanics
A 0.145kg baseball pitched horizontally at 27m/s strikes a bat and pops straight up to a height of 31.5m. If the contact time between the bat and ball is 2.5ms, calculate the average force between the ball and bat during contact.
1438N
2126N
1215N
1566N
2436N
2126N
To figure this out we needed to consider the change in momentum of the ball.
We know that the impulse of the ball is equal to the change in momentum.
The impulse is equal to the force times the time the force is applied.
The change in momentum is equal to the mass times the change in velocity.
Therefore we can combine these equations to say
So let us look at what is happening in the x-direction.
Rearrange and solve the force in the x-direction.
Next, let us determine what is happening in the y-direction. We will need to figure out the initial velocity of the ball in the y-direction using the height. At the peak, we also know the velocity is 0m/s. We also know that the acceleration due to gravity is .
Once we have the initial velocity, we can now determine the force in the y-direction.
Now that we have the force in both the x and y direction we can determine the overall resultant force using the Pythagorean Theorem.