In this video, we’re going to learn about linear momentum. We’ll learn what this term means, how to calculate it, and how it relates to the net force on an object.
To get started, let’s contrast two different scenarios involving object motion. In the first situation, say that you and a friend are playing table tennis. You hit the table tennis ball back and forth and it quickly and easily changes direction. In the second situation, imagine that you’re in your car waiting at a railroad crossing, while a train passes by. The train is made of dozens, if not hundreds, of railroad cars linked together and moves along at a steady speed.
If we compare the motion of the table tennis ball in our first example to the motion of the train, we understand that there is an essential difference between these two. While it is easy to change the direction and speed of the table tennis ball, doing something similar with the train will require quite a lot of force. To understand better the basic difference between these two objects in motion, it will be helpful to learn about linear momentum.
Linear momentum, or momentum for short, describes an object’s capacity to resist changes to its motion. An object’s momentum is equal to its mass multiplied by its velocity. This symbol we usually use to represent momentum is a lowercase 𝑝. And notice that momentum is a vector quantity. Momentum is equal to an object’s mass multiplied by its velocity. And the direction of an object’s momentum is the same as the direction of its velocity.
This mathematical relationship helps us understand the difference between the motion of the table tennis ball and the motion of the train. At any one moment, the magnitude of the velocity of the table tennis ball and the train’s velocity may have been roughly equal. But we know that the mass of the table tennis ball is much, much, much smaller than the mass of the train. So, the train had much more momentum than the table tennis ball and that’s why it’s motion was much harder to change. The more linear momentum an object has, the more force and/or the more time it takes to change that object’s motion.
Now that we have a mathematical relationship for momentum, let’s consider what will happen if we take the time derivative of momentum. 𝑑𝑝 𝑑𝑡 is equal to the time derivative of 𝑚 times 𝑣. And if we apply this operator to the expression 𝑚 times 𝑣 using the chain rule, we find it’s equal to mass times the time rate of change of velocity plus velocity times the time rate of change of mass.
Considering the time rate of change of mass, in a relativistic scenario, the mass of our object may change in time. But outside of that, the mass typically is constant in time so that its time rate of change is zero. When that term is zero, it means that 𝑑𝑝 𝑑𝑡 is equal to the object’s mass times its change in velocity with respect to time. We can recall that 𝑑𝑣 𝑑𝑡 is equal to an object’s acceleration.
Writing the expression this way, it may remind us of Newton’s second law of motion, which says that an object’s mass times its acceleration is equal to the net force acting on the object. So, an object’s time rate of change in momentum is equal to the net force acting on it. And if we multiply both sides of this equation by 𝑑𝑡, we find that 𝑑𝑝, the change in momentum, is equal to net force times the change in time, or in a notation that may be more familiar, Δ𝑝 is equal to 𝐹 net times Δ𝑡.
Now, we not only understand momentum as a product of mass and velocity, but we see that an object’s change in momentum is equal to the net force acting on it times the time over which that force acts. Just as a side note, this relationship as it’s written assumes that our net force is constant in time. Now that we know a bit about momentum and how it can change, let’s get some practice with these ideas through a couple of examples.
A hockey puck of mass 150 grams is sliding due east on a frictionless table with a speed of 10 meters per second. Suddenly, a constant force of magnitude 5.0 newtons and direction due north is applied to the puck for 1.5 seconds. Find the northward component of the hockey puck’s momentum at the end of the 1.5-second interval. Find the eastward component of the hockey puck’s momentum at the end of the 1.5-second interval.
We can refer to the northward component of the puck’s momentum and the eastward component of the puck’s momentum at the end of this interval as 𝑝 sub 𝑛 and 𝑝 sub 𝑒, respectively. So, we want to solve for 𝑝 sub 𝑛 and 𝑝 sub 𝑒, the northward and eastward components of the puck’s momentum , after a time, we’ve called Δ𝑡, of 1.5 seconds has elapsed.
Before this time interval begins, we’re told that the puck is moving to the east with a speed of 10 meters per second. At time 𝑡 equals zero, a force 𝐹 is applied to the puck and lasts 1.5 seconds. Regarding momentum, there are two relationships we can recall. The first is that an object’s momentum is equal to its mass times its velocity. And the second is that the change in an object’s momentum is equal to the net force acting on it times the time over which that force acts.
Considering the northward component of our puck’s momentum, we know that at the outset, it has no momentum in that direction because its velocity to the north is zero. But after 1.5 seconds of the force 𝐹 being applied pointing to the north, referring to our second relationship for momentum, we can write that 𝑝 sub 𝑛 is equal to the magnitude of 𝐹 multiplied by Δ𝑡.
This change in momentum to the north is equal to the final northward momentum component because the original component was zero. Plugging in for these two values, we find that 𝑝 sub 𝑛 is equal to 7.5 kilograms meters per second. That’s the northward component of the puck’s momentum after a time Δ𝑡 has passed.
Next, we wanna solve for the eastward component of the momentum after that same time interval. Unlike to the north, to the east our puck does have an initial speed. And because the force applied to the puck is perpendicular to that direction, it doesn’t affect the puck’s speed to the east. This means the puck’s momentum to the east is a constant value. It doesn’t change in time and is always equal to the puck’s mass times 𝑣 sub 𝑖.
We’re told that the mass of the puck is 150 grams. But when we enter it into our expression, we write it in units of kilograms to be consistent with the units in the rest of the expression. Calculating this product, we find that 𝑝 sub 𝑒 is 1.5 kilograms meters per second. That’s the puck’s eastward momentum component after Δ𝑡 has passed or at any time. Next, let’s look at an example involving momentum and projectile motion.
A tennis ball of mass 150 grams is thrown at an angle of 40 degrees above the horizontal and a speed of 18 meters per second. What is the momentum of the ball after it has been in flight for 0.36 seconds?
We can call the ball’s momentum after it’s been in flight for this amount of time 𝑝 and begin on our solution by drawing a diagram of this situation. Our sketch represents that we’ve thrown this ball at an angle, we’ve called 𝜃, of 40 degrees above the horizontal at an initial speed, we’ve called 𝑣 sub 𝑖, of 18 meters per second. After a time in flight, we’ve called Δ𝑡, of 0.36 seconds, we want to know the momentum of the ball.
If we let upward motion be motion in the positive 𝑦-direction and motion to the right be in the positive 𝑥 and if we recall that momentum 𝑝 is equal to an object’s mass times its velocity, then we can write that momentum 𝑝 is equal to our ball’s mass 𝑚 multiplied by the square root of 𝑣 sub 𝑥 squared plus 𝑣 sub 𝑦 squared. Here, 𝑣 sub 𝑥 is the ball’s speed in the 𝑥-direction at a time value of 0.36 seconds. And likewise, 𝑣 sub 𝑦 is the ball’s vertical speed at that same time.
Since we’ve been told the ball’s mass 𝑚, that means if we can solve for 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we’ll be able to solve for momentum 𝑝. Starting with 𝑣 sub 𝑥, when we look at our diagram, we see that we can represent the speed of the ball in the 𝑥-direction in terms of 𝑣 sub 𝑖 and 𝜃. 𝑣 sub 𝑥 which, recall, is the ball’s 𝑥-velocity component at time 𝑡 equals 0.36 seconds is equal to the ball’s initial launch speed 𝑣 sub 𝑖 times the cosine of the angle 𝜃.
Plugging in for these values, when we calculate 𝑣 sub 𝑥, we find it’s approximately 13.79 meters per second. This is the ball’s speed in the 𝑥-direction and is constant throughout the ball’s flight. Now that we know 𝑣 sub 𝑥, we move on to solving for 𝑣 sub 𝑦, the ball’s vertical speed component at time 𝑡 equal 0.36 seconds.
Unlike the ball’s 𝑥-component speed, its 𝑦-component speed is not constant in time, but constantly changes under the influence of gravity. As a projectile, the ball’s motion can be described by the kinematic equations. In particular, we recall the relationship final speed is equal to initial speed plus acceleration times time. When we write this equation in terms of our variables, we can write that 𝑣 sub 𝑦 is equal to the initial speed of the ball in the 𝑦-direction minus the acceleration due to gravity 𝑔 times Δ𝑡.
We’ll treat gravitational acceleration 𝑔 as exactly 9.8 meters per second squared. Looking again at our diagram, we see that we can write the expression for 𝑣 sub 𝑦𝑖, the initial speed of the ball in the vertical direction, as the product of 𝑣 sub 𝑖, the ball’s initial speed, times the sin of the angle 𝜃. Since we’re given 𝑣 sub 𝑖, 𝜃, and Δ𝑡, and 𝑔 is a constant, we’re ready to plug in and solve for 𝑣 sub 𝑦.
When we do and enter this expression on our calculator, we find that the 𝑣 sub 𝑦 is approximately 8.042 meters per second. Now that we know 𝑣 sub 𝑥 and 𝑣 sub 𝑦, we’re ready to plug in and solve for momentum 𝑝. When we do plug in, we’re careful to convert our mass into units of kilograms. And we find that 𝑝 is equal to 2.4 kilograms meters per second. That’s the overall momentum of this ball 0.36 seconds after its been launched.
Let’s summarize what we’ve learned about linear momentum. We’ve seen that an object’s momentum describes its capacity to resist changes to its motion. The motion of our table tennis ball was easy to change and it had a small momentum, while our train’s momentum was very large indicating that its motion was difficult to change. We’ve also seen that an object’s momentum is equal to its mass multiplied by its velocity.
Mathematically, we write that momentum, symbolized by a lowercase 𝑝, is equal to 𝑚 times 𝑣. And we’ve noted that momentum is a vector quantity with magnitude as well as direction. And finally, we consider that the change in an object’s momentum Δ𝑝 is equal to the net force acting on that object times the time Δ𝑡 over which that force acts. Linear momentum is a helpful tool for understanding differences in object motion.