Impulse and Momentum - AP Physics 1
Card 1 of 340
One car with a mass of 400kg is traveling east at 
and collides with a car of mass 800kg traveling west at 
. Assuming the collision is completely inellastic, what is velocity of the first car after the collision?
One car with a mass of 400kg is traveling east at
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Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:
There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:
Rearranging for final velocity, we get:
At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:
Since this value is positive, the final answer is 
West.
Since the collision is completely inelastic, momentum is conserved but energy is not. Furthermore, the two cars stick to each other and travel as one. The equation for conservation of momentum is as follows:
There are two inital masses with different velocities and one final mass with a single velocity. Therefore, we can write:
Rearranging for final velocity, we get:
At this point, we can denote which direction is positive and which is negative. Since the car traveling west has more momentum, we will consider west to be positive. Substituting our values into the equation, we get:
Since this value is positive, the final answer is
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A popular topic in early space exploration was how to safely return modules back to the surface of the earth. Early designs contained materials that could only withstand impulses of up to 
before parts of the module became compromised. One such model of mass 500kg is approaching ocean waters and deploys its shoot, reducing its speed to 
. If the module decelerates to zero velocity in 0.9 seconds upon hitting the water, is the module structurally compromised?
A popular topic in early space exploration was how to safely return modules back to the surface of the earth. Early designs contained materials that could only withstand impulses of up to
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We need to use the equation for impulse to solve this problem. In fact, the time given is completely irrevelevant:
Plugging in our values:
This is less than the threshold, so no, nothing becomes structurally compromised
We need to use the equation for impulse to solve this problem. In fact, the time given is completely irrevelevant:
Plugging in our values:
This is less than the threshold, so no, nothing becomes structurally compromised
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Two sumo wrestlers are in a match. At the start of the match, they both lunge at each other. They hit and miraculously come to a stand still. One wrestler was 200kg and traveling at a velocity of 
at the instance of collision. If the other wrestler was traveling at 
, what is his mass?
Two sumo wrestlers are in a match. At the start of the match, they both lunge at each other. They hit and miraculously come to a stand still. One wrestler was 200kg and traveling at a velocity of
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It does not matter whether the collision is elastic or inelastic (although it would be best to assume that it's inelastic). Momentum is conserved in either type of collision, and is the only value needed for our calculation. Since they come to a standstill, their momentums at the moment of collision are equal and opposite:
Rearrange to solve for 
:
Plug in the given values from the question and solve:
It does not matter whether the collision is elastic or inelastic (although it would be best to assume that it's inelastic). Momentum is conserved in either type of collision, and is the only value needed for our calculation. Since they come to a standstill, their momentums at the moment of collision are equal and opposite:
Rearrange to solve for
Plug in the given values from the question and solve:
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In a billiards game, one player hits the cue ball towards another ball. The cue ball has a mass of 0.1kg and hits the other ball with a velocity of 
. If the collision is completely elastic and the cue ball travels with a velocity of 
after the collision, what is the mass and velocity of the other ball?
In a billiards game, one player hits the cue ball towards another ball. The cue ball has a mass of 0.1kg and hits the other ball with a velocity of
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Since the collision is completely elastic, we know that both momentum and kinetic energy are conserved. We can write the following equations (initial momentum and energy of the second ball are neglected since it is not moving:
Rearrange the first equation for 
and the second for 
.
We can rewrite the second equation as:
Substitute our equation for 
into the second equation:
Rearranging, we get:
Plug in our values for the initial and final velocities:
To solve for mass, we'll use our earlier expression for 
:
Since the collision is completely elastic, we know that both momentum and kinetic energy are conserved. We can write the following equations (initial momentum and energy of the second ball are neglected since it is not moving:
Rearrange the first equation for
We can rewrite the second equation as:
Substitute our equation for
Rearranging, we get:
Plug in our values for the initial and final velocities:
To solve for mass, we'll use our earlier expression for
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A marble of mass 
falls off a bed with a height of 
. What is the impulse on the marble as it hits the ground?
A marble of mass
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Impulse can be written as either of two popular expressions:
From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:
Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:
Canceling out mass and rearranging for final velocity, we get:
We know these variables, allowing us to solve for the velocity:
Plugging this value into the expression for impulse, we get:
Impulse can be written as either of two popular expressions:
From the problem statement, we can determine the velocity of the marble as it hits the floor, allowing us to use the latter expression. To determining the velocity of the marble, we can use the equation for conservation of energy:
Assuming the final height is zero, we can eliminate initial kinetic energy and final potential energy. Therefore, we can write:
Canceling out mass and rearranging for final velocity, we get:
We know these variables, allowing us to solve for the velocity:
Plugging this value into the expression for impulse, we get:
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Consider the following system:
If the block has a mass of 
, the angle measures 
, and there is no friction between the block and slope, what is the momentum of the block after it has traveled a horizontal distance of 
?
Consider the following system:
If the block has a mass of
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To calculate the momentum of the block, we first need to know the velocity of the block. This can be found using the equation for the conservation of momentum:
If we assume that the final height is zero, we can eliminate initial kinetic energy and final potential energy, getting:
Substituting expressions for each term, we get:
Cancel out mass and rearrange to solve for velocity:
We can use the horizontal distance traveled and the angle of the slope to determine the initial height:
Now that we have the initial height, we can solve for final velocity:
Finally, we can now use the equation for momentum to solve the problem:
To calculate the momentum of the block, we first need to know the velocity of the block. This can be found using the equation for the conservation of momentum:
If we assume that the final height is zero, we can eliminate initial kinetic energy and final potential energy, getting:
Substituting expressions for each term, we get:
Cancel out mass and rearrange to solve for velocity:
We can use the horizontal distance traveled and the angle of the slope to determine the initial height:
Now that we have the initial height, we can solve for final velocity:
Finally, we can now use the equation for momentum to solve the problem:
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Two astronauts in space are traveling directly towards each other. Astronaut A has a mass of 
and a velocity of 
and Astronaut B has a mass of 
and a velocity of 
. When the astronauts collide, they grab onto each other. What is the velocity of the two astronauts after the collision as they continue to grab onto each other?
Two astronauts in space are traveling directly towards each other. Astronaut A has a mass of
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Momentum is always conserved. Equation for conservation of momentum:
There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.
Momentum is always conserved. Equation for conservation of momentum:
There is only one velocity on the right since the two astronauts grab onto each other, thus they move together at the same velocity. Solve.
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Tom drops a ball of mass 
from rest from a height 
. The ball bounces back to a height of 
. Find the magnitude of the impulse the ground imparted on the ball.
Tom drops a ball of mass
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Impulse is just the change in momentum. To find the velocity when the ball hits the ground, we need to use kinematics. We know the height the ball is dropped, the acceleration, and the initial velocity, so we can use the equation 
. The initial velocity is 
, 
, and 
, so the equation becomes 
When the ball bounces back up it reaches a height of 
. In order to find the velocity immediately after it hits the ground, we can use the same equation with 
. This will lead it a velocity of 
Assuming up is positive, the magnitude of the impulse is just 
Impulse is just the change in momentum. To find the velocity when the ball hits the ground, we need to use kinematics. We know the height the ball is dropped, the acceleration, and the initial velocity, so we can use the equation
When the ball bounces back up it reaches a height of
Assuming up is positive, the magnitude of the impulse is just
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Joe, of mass 90kg, jumps straight up. To do so, he bends his knees and produces an upwards force that results in a constant upward net force of 100N. If Joe experiences this force for 0.9s before leaving the ground, what is Joe's velocity immediately after he leaves the ground?
Joe, of mass 90kg, jumps straight up. To do so, he bends his knees and produces an upwards force that results in a constant upward net force of 100N. If Joe experiences this force for 0.9s before leaving the ground, what is Joe's velocity immediately after he leaves the ground?
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To solve this problem we need to use the relationship between force and impulse, which is given by the following equation:
This equation represents that the rate of change of momentum with respect to time is equal to the net force that causes said change in momentum. Thus:
Note that Joe must have an initial velocity of 
before he begins to apply the upwards Force that accelerates him upwards, therefore our equation simplifies to:
Solve for 
:
To solve this problem we need to use the relationship between force and impulse, which is given by the following equation:
This equation represents that the rate of change of momentum with respect to time is equal to the net force that causes said change in momentum. Thus:
Note that Joe must have an initial velocity of
Solve for
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Which of the following explains why when we land on our feet, we instinctively bend our knees? Hint: think about the relationship between force, impulse, and time.
Which of the following explains why when we land on our feet, we instinctively bend our knees? Hint: think about the relationship between force, impulse, and time.
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Say that, when we hit the ground, we have a velocity 
, which is predetermined by whatever happens before the impact. When we hit the ground you will experience a force for some time. This force will cause the acceleration that reduces our velocity to zero and gets us to stop. Note that, regardless of how much time it takes us to stop, the change in momentum (impulse) is fixed, since it directly depends on how much our velocity changes:
(since we come to a stop)
Note that the initial momentum does not depend on the impact force nor on how much time it takes to stop. The initial momentum depends on the velocity we have when we first hit the ground. This velocity is given by whatever happened before we hit the ground, which no longer concerns us since we only care about what happens from the moment we first hit the ground till the moment we stop. Yes, the time that passes for you to stop is very small, but it is impossible for it to be zero. So we have that the change in momentum (impulse) is a constant:
, since 
is predetermined.
Remember that any change in momentum for a given mass occurs because its velocity changes. The velocity of the mass changes due to an acceleration and an acceleration is caused by a force. This gives us a relationship between force and impulse:
In our scenario, 
would be the impact force that stops us and 
the time it takes us to stop. From the equation above, it is easy to see that, since 
is fixed, when 
gets larger 
gets smaller, and the other way around. Therefore, we bend our knees to effectively increase the time it takes us to stop. Thus, diminishing the impact force as to avoid hurting ourselves.
Say that, when we hit the ground, we have a velocity
Note that the initial momentum does not depend on the impact force nor on how much time it takes to stop. The initial momentum depends on the velocity we have when we first hit the ground. This velocity is given by whatever happened before we hit the ground, which no longer concerns us since we only care about what happens from the moment we first hit the ground till the moment we stop. Yes, the time that passes for you to stop is very small, but it is impossible for it to be zero. So we have that the change in momentum (impulse) is a constant:
Remember that any change in momentum for a given mass occurs because its velocity changes. The velocity of the mass changes due to an acceleration and an acceleration is caused by a force. This gives us a relationship between force and impulse:
In our scenario,
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When catching an object, an average person can stand a maximum impact force of 20000N. Forces greater than this would most likely break bones in the person's hand. If a person catches a 500g baseball that moves at 
, what is the minimum time the person should take to stop it in order to avoid seriously hurting his hand?
When catching an object, an average person can stand a maximum impact force of 20000N. Forces greater than this would most likely break bones in the person's hand. If a person catches a 500g baseball that moves at
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In order to solve this problem we need to use the relationship between force and impulse:
Since the ball is moving with a Velocity of 
, we have that
Note that the final velocity of the ball is zero since it comes to a stop. We want the force, 
, experienced by the person's hand to be less than or equal to the maximum impact force
Mathematically:
Use the impulse equation and solve for time:
In order to solve this problem we need to use the relationship between force and impulse:
Since the ball is moving with a Velocity of
Note that the final velocity of the ball is zero since it comes to a stop. We want the force,
Mathematically:
Use the impulse equation and solve for time:
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A 150g baseball is thrown with a speed of 
. If it takes 0.7s for the baseball to come to rest in the catcher's glove, what is the average force the catcher experiences due to the ball?
A 150g baseball is thrown with a speed of
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To solve this problem, we need to consider the change in the ball's momentum. To do so, we'll use the following equation.
Rearrange the above equation to solve for the average force.
To solve this problem, we need to consider the change in the ball's momentum. To do so, we'll use the following equation.
Rearrange the above equation to solve for the average force.
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A 
rock flying through the air is traveling at a velocity of 
when it collides into and sticks to a stationary bean bag, weighing 
. What is the velocity of the two objects?
A
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The equation for momentum is:
To maintain conservation of momentum, a new state must have the same momentum as a previous state:
Since the rock and the bean bag move together after the collision, 
. And since the bean bag is initially stationary, 
Plug in known values and solve.
The equation for momentum is:
To maintain conservation of momentum, a new state must have the same momentum as a previous state:
Since the rock and the bean bag move together after the collision,
Plug in known values and solve.
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An asteroid of mass 
is traveling with the velocity 
.
What is the magnitude of the momentum of the asteriod?
An asteroid of mass
What is the magnitude of the momentum of the asteriod?
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First, we will need to find the magnitude of the velocity vector.
Plugging in our values
Momemtum is defined as
Thus,
We plug in our values
First, we will need to find the magnitude of the velocity vector.
Plugging in our values
Momemtum is defined as
Thus,
We plug in our values
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A train of mass 
traveling at 
strikes a car stuck on the tracks of mass 
.
Determine the initial momentum of the system.
A train of mass
Determine the initial momentum of the system.
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The train and car are our only two objects in the system.
The initial momentum of the car is zero.
So the only momentum that will contribute is that of the train.
Plugging in our values, we get 
The train and car are our only two objects in the system.
The initial momentum of the car is zero.
So the only momentum that will contribute is that of the train.
Plugging in our values, we get
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A train of mass 
traveling at strikes a car stuck on the tracks of mass . The car becomes stuck on the train.
Determine the final velocity of the train.
A train of mass
Determine the final velocity of the train.
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We will need to use conservation of momentum to solve this problem.
Where 
and 
refer to the train, and 
and 
refer to the car. 
and 
refer to the final state of both the train and the car.
Rearranging using algebra......
Plugging in our values, we get 
.
We will need to use conservation of momentum to solve this problem.
Where
Rearranging using algebra......
Plugging in our values, we get
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A train of mass 
traveling at strikes a car stuck on the tracks of mass . The car becomes stuck on the train.
Determine the final velocity of the car.
A train of mass
Determine the final velocity of the car.
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We will need to use conservation of momentum to solve this problem.
Where and refer to the train, and and refer to the car. and refer to the final state of both the train and the car.
Rearranging using algebra......
Plugging in our values, we get .
We will need to use conservation of momentum to solve this problem.
Where
Rearranging using algebra......
Plugging in our values, we get
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A train of mass 
traveling at 
strikes a car stuck on the tracks of mass 
.
Let's assume this collison took 
to happen. That is, it took the car 
to accelerate to it's new velocity. Determine the force experienced by the car.
A train of mass
Let's assume this collison took
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We will need to use conservation of momentum to solve this problem.
Where and refer to the train, and and refer to the car. and refer to the final state of both the train and the car.
Rearranging using algebra......
Plugging in our values, we get .
Then, we will need to find out final momentum of the car.
Since our intial momentum of the car was 
, our change in momentum will be equal to the 
.
We will use the definition of impulse, which is the change in momentum:
We will use substitution:
Plugging in our values, we get 
We will need to use conservation of momentum to solve this problem.
Where
Rearranging using algebra......
Plugging in our values, we get
Then, we will need to find out final momentum of the car.
Since our intial momentum of the car was
We will use the definition of impulse, which is the change in momentum:
We will use substitution:
Plugging in our values, we get
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Imagine a baseball player hitting a home run. If the 1 kg ball is thrown at 
and it leaves the bat at 
. What is the impluse applied by the bat to the ball?
Assume the collision lasts 
of a second and the ball leaves at the same angle it entered.
Imagine a baseball player hitting a home run. If the 1 kg ball is thrown at
and it leaves the bat at
Assume the collision lasts
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Impulse is the change in momentum. So all that is needed for this problem is to solve for the change in momentum.
Note!!!! momentum is a vector quantity. This means that we must accound for the change in the ball's direction. This can be done by defining one of the directions negative.
For convience, I'll define the initial velocity as negative. Plugging in numbers we get
.
We never accounted for the time of the collision. We would have needed to do this if the problem asked for the force the bat applied.
Impulse is the change in momentum. So all that is needed for this problem is to solve for the change in momentum.
Note!!!! momentum is a vector quantity. This means that we must accound for the change in the ball's direction. This can be done by defining one of the directions negative.
For convience, I'll define the initial velocity as negative. Plugging in numbers we get
We never accounted for the time of the collision. We would have needed to do this if the problem asked for the force the bat applied.
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A 
bullet is fired at a 
block of lead resting on a frictionless surface. The bullet has an initial speed of 
, while the block is initially at rest. After hitting the block, the bullet rebounds with a speed of 
. How fast is the lead block moving after the bullet rebounds off of it?
A
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To solve this problem, we will use conservation of momentum. The initial momentum of the system must be equal to the final momentum of the system if no external forces act on it. It is important to note the directions and signs of the velocities. From this information, we may write:
To solve this problem, we will use conservation of momentum. The initial momentum of the system must be equal to the final momentum of the system if no external forces act on it. It is important to note the directions and signs of the velocities. From this information, we may write:
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