11.1.8.1: Illustrations

Illustration 1: Force and Impulse

So what do we mean by a force? Newton considered a net force as something that caused a time rate of change of momentum, \(\Delta \mathbf{p}/\Delta t\) or \(d\mathbf{p}/dt\). However, C.D. Broad (Scientific Thought, 1923) wrote, "It seems clear to me that no one ever does mean or ever has meant by 'force', rate of change of momentum." So if Newton's statement seems odd it is because you are used to a special—and famous—case of Newton's general statement of the second law, that of \(\sum\mathbf{F}_{\text{net}}=m\mathbf{a}\). Restart.

Consider the force applied by the hand over a small \(\Delta t\)  (this happens automatically at \(t = 1\text{ s}\)). Notice the change in momentum (position is given in meters and time is given in seconds). The arrow represents the change in momentum. Initially the mass of the cart is \(1\text{ kg}\). Change the mass to \(2\text{ kg}\). Does the change in momentum differ? No! But what does change is the final velocity; it is half of the velocity when the mass was \(1\text{ kg}\). The same force results in the same change in momentum in the same time interval.

Another way to represent this is in terms of the integral (the area) under a force vs. time graph. Check the box to see this graph. This area is called the impulse, which is a fancy name for \(\Delta\mathbf{p}\). What can you say about the impulse received by the cart, independent of its mass? Check the second box to find out. Again, it should be, and is, the same.

Consider the animation with the force applied by the hand over a large \(\Delta t\)  (this happens automatically at \(t = 1\text{ s}\)). The difference between the animations is that in large \(\Delta t\) the force acts for a longer time and therefore the force causes a larger change in momentum. Again, the arrow represents the change in momentum, which is larger than the small \(\Delta t\) case.

Illustration 2: The Difference between Impulse and Work

In Illustration 8.1, we learned a change in momentum was due to a net applied force. What about kinetic energy? Well, it is also due to a net applied force, but in a different way. Recall that we talk about work as the amount of force in the direction of an object's displacement multiplied by the displacement. No displacement, no work. Work is positive if \(\mathbf{F}\) and \(\Delta\mathbf{x}\) (or \(d\mathbf{x}\)) point in the same direction and negative if \(\mathbf{F}\) and \(\Delta\mathbf{x}\) (or \(d\mathbf{x}\)) point in opposite directions. Restart.

Consider the force applied by the hand over a small \(\Delta t\) (this happens automatically at \(t = 1\text{ s}\)). Notice the change in momentum (position is given in meters and time is given in seconds). Initially the mass of the cart is \(1\text{ kg}\). Change the mass to \(2\text{ kg}\). Does the change in momentum differ? No!  But what does change is the final velocity; it is half of the velocity when the mass was \(1\text{ kg}\). The same force results in the same change in momentum in the same time interval.

So what happens to the kinetic energy? Does it remain the same upon a change in mass? No. Why not? Recall that the work that will be equal to the change in kinetic energy is related to the displacement the cart undergoes when the force is applied. Due to the larger mass, the cart does not accelerate as much and therefore does not move as far, so its kinetic energy is less.

Another way to represent this is in terms of the integral (the area) under a force \(\cos(\theta )\) vs. distance graph. This area is called the work that is the object's \(\Delta KE\). What can you say about the work received by the cart when its mass changes? Check the second box to find out. Again, it should be, and is, different.

What happens when instead of applying a small \(\Delta t\), you apply a large \(\Delta t\)? There is a larger impulse because \(\Delta t\) is larger. There is also a larger change in the kinetic energy since \(\Delta x\) is larger as well.

Illustration 3: Hard and Soft Collisions and the Third Law

This Illustration models collisions between two particles (position is given in meters and time is given in seconds). Restart. Both animations assume identical particles and both animations start with the particles having the same velocities. What is different about the animations is the interaction between the two particles. The interaction in Animation 1 can be characterized as hard since the acceleration is very large and the interaction is very short range. It is also called a contact interaction because the force turns on when the two particles are in contact.

In Animation 2 the interaction can be characterized as soft. Vary the masses of the two particles and be sure to pay attention to the scale of the acceleration graph in each animation. In addition, watch the relative acceleration of the particles as you vary the masses. Notice that the accelerations are different when the masses are different.

What can you say about the force experienced by each particle in each animation? The character of the forces is different: One is soft and one is hard. Nonetheless, the forces are always equal and opposite. To see this you must take each object's acceleration and multiply it by its mass. This is exactly the statement of Newton's third law, the law of reciprocity of forces. In this case, there are no net external forces acting on the two particles, so the change in momentum of the two-particle system is zero. In other words, momentum is conserved. Using equations, we would say that since \(\sum\mathbf{F}_{\text{net}}=\Delta\mathbf{P}/\Delta t\) or \(\sum\mathbf{F}_{\text{net}}=d\mathbf{p}/dt\), if the net force on a system is zero, then \(\Delta\mathbf{p}/\Delta t=0\) or \(d\mathbf{p}/dt = 0\), which means that the change in momentum over time must be zero. Hence the sum of the two impulses experienced by the balls must be zero. If one particle's momentum goes up, the other particle's momentum must go down by exactly the same amount. Check it out by looking at the tables.

Two-dimensional models show a dramatic difference between hard and soft collisions. (See Problem 8.12 for two-dimensional collisions.) Hard collisions tend not to have much of an effect on incident particles except for the occasional particle that suffers a head-on impact. Soft collisions, on the other hand, produce minor deflections on a large number of particles. The experimental observation of alpha particles being deflected backwards from gold foil led Ernest Rutherford to predict that atoms have a small hard core, the nucleus.

Illustration 4: Relative Velocity in Collisions

In this set of collisions there are no net external forces acting on the two carts. Restart. Enter new values for the velocity of each cart and the mass of the right-moving (orange) cart. Then click the "set values and play" button to register your values and play the animation (position is given in meters and time is given in seconds). We have set limits on the values you can choose:

\[0.5\text{ kg} < m_{1} < 4\text{ kg},\quad 0\text{ m/s} < v_{1} < 4\text{ m/s},\quad\text{ and}\quad -4\text{ m/s} < v_{2} < 0\text{ m/s.}\nonumber\]

The table gives an instantaneous reading of each cart's momentum as well as the total momentum in the two-cart system. In addition, when you select the check box, arrows representing the magnitude of the relative velocities before and after the collision between the two carts are also shown.

Because the net force on the system of two carts is zero, the change in momentum of the two-particle system is zero. In other words, momentum is conserved. Using equations, we would say that since \(\sum\mathbf{F}_{\text{net}}=\Delta\mathbf{p}/\Delta t\) or \(\sum\mathbf{F}_{\text{net}}=d\mathbf{p}/dt\), if the net force on a system is zero, then \(\Delta\mathbf{p}/\Delta t = 0\) or \(d\mathbf{p}/dt = 0\), which means that the change in momentum over time must be zero. Hence the sum of the two impulses experienced by the carts must be zero. If one particle's momentum goes up, the other particle's momentum must go down by exactly the same amount.

In elastic collisions, the concept of the relative velocity is an important one in analyzing the collision. The relative velocity is defined as \(\mathbf{v}_{1}-\mathbf{v}_{2}\) (it could also be defined as \(\mathbf{v}_{2}-\mathbf{v}_{1}\) as the choice of \(1\) and \(2\) is arbitrary).

Turn on the relative velocity arrows and vary the velocity of each cart and the mass of the right-moving (orange) cart. Determine the relationship between the relative velocity before the collision and the relative velocity after the collision. What did you find? It turns out that the magnitude of the relative velocity before and after an elastic collision is the same. However, the sign of the relative velocity changes from before to after the collision: \(\mathbf{(v}_{1}-\mathbf{v}_{2}\mathbf{)}_{\mathbf{i}}=\mathbf{-(v}_{1}-\mathbf{v}_{2}\mathbf{)}_{\mathbf{f}}\). This relationship can be verified by using the conservation of energy and conservation of momentum equations and a bit of algebra.

Consider an elastic collision where \(v_{1}= 1\text{ m/s}\) and \(v_{2} = -4\text{ m/s}\). Clearly the relative velocity before the collision is \(5\text{ m/s}\). What must it be after the collision? \(-5\text{ m/s}\). Try it and find out if this is true. Does it matter if you change the mass of the orange cart?

Exploration 5: The Zero-Momentum Frame

Is physics different when viewed in different reference frames? Well, it can certainly look different. Consider the collision in the animation as seen initially in the reference frame of Earth (the relative velocity between this frame and Earth's stationary frame is zero). Here both the red ball and the blue ball have the same mass equal to \(1\text{ kg}\). Note that energy and momentum are conserved in the collision with \(KE = 2\text{ J}\) and \(p_{x} = 2\text{ kg}\cdot\text{m/s}\) before and after the collision. Restart.

Change \(v\) from zero to \(2\text{ m/s}\) (position is given in meters and time is given in seconds). How does the collision change? The red ball is now initially stationary, and the blue ball is moving to the left at \(2\text{ m/s}\). Note that in the original collision with \(v = 0\text{ m/s}\), the red ball was initially moving to the right and the blue ball was initially stationary. In the new frame the momentum of the two-ball system is different. However, the kinetic energy happens to be the same and energy and momentum are conserved.

Now try \(v = -2\text{ m/s}\). Are energy and momentum still conserved? Even though the values of the kinetic energy and momentum change, the laws of conservation of energy and conservation of momentum still hold.

Now try \(v = 1\text{ m/s}\). What is the new momentum for the two-ball system? This frame of reference is appropriately called the zero-momentum frame. In this frame the sum of the momentum of all objects in the system is zero. This frame is also called the center-of-mass frame. The center of mass is a coordinate that is a mass-weighted average of the positions of the objects that make up the system. In a two-object system the center of mass is always somewhere in between the two objects. Since the center of mass is a mass-weighted average, the center of mass will always be closer to the object that is more massive. In the case of this animation, where both balls have the same mass, the center of mass is always at the midpoint between the two masses. This point does not move in the zero-momentum frame, but does move in other frames.

Illustration 6: Microscopic View of a Collision

In the animation, a red \(80\text{-kg}\) ball with an initial kinetic energy of \(360\text{ J}\) is trapped inside a box with rigid walls containing a cylinder constructed of \(80\) small \(1\text{-kg}\) spheres (position is given in meters, time is given in seconds, and energy shown on the bar graph is given in joules). The ball crashes into the cylinder and breaks it apart. The bar graph displays the kinetic energy of the red ball. The table displays the time, momentum, and kinetic energy of the red ball. Restart.

This animation is meant to simulate a collision between two solid objects, one of which is stationary. The stationary object is a loose collection of smaller objects and approximates a larger solid object. This is only an approximation since this object should stay together, not break apart. When you do collision experiments in the lab the objects colliding do not usually deform this much! Nevertheless, we can learn a lot from this animation. As the red ball hits the blue object, the blue object deforms, absorbing some of the red ball's kinetic energy and momentum. If the blue object were indeed solid, the deformed object—the entire object—would move to the right. We can imagine this by considering the average motion of the small blue balls that make up the larger solid object. We note that the general motion of these blue balls is to the right. Where does all of the initial kinetic energy go? It goes into the kinetic energy of the small blue balls.

Illustration 7: Center of Mass

The first image shows two blocks of equal mass (position is given in meters). The center of mass of the system is shown by a red dot and its position is calculated for you. Drag the green block on the right. What do you notice about the location of the center of mass as you change the right block's position? Restart.

Now suppose the two blocks have unequal mass as shown in Animation 2. Is the center of mass at the center of the system? By viewing the location of the center of mass, you know which block is more massive. So which one is more massive?

How can we calculate the ratio of the mass of the blue block to the mass of the red block? If you only have two blocks, then the ratio of the distances of each block from the center of mass is related to the ratio of their masses. Therefore, by measuring the distance from each block to the center of mass, you can calculate the ratio of their masses. The location of the center of mass is given by \(X_{\text{cm}} = (x_{1}m_{1} + x_{2}m_{2})/(m_{1} + m_{2})\) for a one-dimensional two-object system.

A concept similar to that of the center of mass is that of the center of gravity. In fact the two are often used interchangeably. The center of mass is defined above; the center of gravity is defined as the point on a system where gravity can be considered to act. The center of gravity takes into account the fact that the force of gravity—and therefore the acceleration due to gravity—is different for different heights above the surface of Earth. For this Illustration, the center of mass is equivalent to the center of gravity. Only if the system is really large might the acceleration due to gravity be different at different parts of the system. This would cause the center of gravity to differ from the center of mass.

Illustration authored by Aaron Titus and Mario Belloni.
Script by Aaron Titus.

Illustration 8: Moving Objects and Center of Mass

A green block, \(1.00\text{ kg}\), sits on a red block, \(4.46\text{ kg}\), as shown in the animation (position is given in meters and time is given in seconds). Restart. All surfaces are frictionless except for the gray patches on the red block. Given the self-propelled motion of the green block in Animation 1, are momentum and energy conserved in the animation? If not, why not?

Well, momentum is conserved because there are no external forces. The momentum of the system was zero before the green block moved, is zero when the blocks move, and is again zero when the blocks are stationary. From the point of view of the center of mass, \(V_{\text{cm}} = 0\text{ m/s}\) and therefore \(P_{\text{cm}} = 0\text{ kg}\cdot\text{m/s}\).

We can see this by considering what happens to the center of mass during Animation 2. The center of mass of the system is \(X_{\text{cm}} = (m_{1}x_{1} + m_{2}x_{2})/(m_{1} + m_{2})\) and is represented by the black dot. Note that the center of mass of the system does not move relative to the ground but does move to the left relative to the right edge of the red block as the red block moves to the right. In fact we can look at the center of mass for each object by replacing each block by a dot as well as shown in Animation 3.

What about energy? As is always the case, whether energy is conserved depends on how you define your system. Looking just at the center of mass, since \(V_{\text{cm}} = 0\text{ m/s}\), energy is conserved. However, if we look at the blocks individually, energy (in the sense of mechanical energy) is not conserved. Energy stored in the individual elements of the system (presumably the green block's potential energy) is turned into kinetic energy of both blocks and then is dissipated by friction.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.