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Example Questions
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.
Example Question #291 : Motion And Mechanics
A 950kg sports car collides into the rear end of 2200kg SUV stopped at a red light. The bumper lock, the brakes lock, and the two cars skid forward 3.0m before coming to a stop. The police officer, knowing the coefficient of kinetic friction between the tires and the road is about 0.8. He calculates the speed of the sports car at impact. What was that speed?
46.7m/s
6.9m/s
22.9m/s
17.4m/s
20.5m/s
22.9m/s
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 at 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 the 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 #292 : Motion And Mechanics
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 550kg and the car behind it (CarB) has a mass of 450kg. The car in front was traveling at 3.70m/s while the car behind hit him with a velocity of 4.50m/s. 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 let's substitute in values and solve.
We can now plug this value back into our equation for