Momentum and Its Conservation

Momentum Newton defined momentum as the quantity of motion. It is the product of an object's mass and motion. Use the equation p = mv, p in kg*m/s.

Q. What is the momentum of a 200 kg cow lumbering along at 0.2 m/s?

p = mv so p = 200 kg * 0.2 m/s = 40 kg*m/s.

Newton's first Law of Motion can be restated as: with no net force acting on an object, its momentum does not change, i.e., the object's momentum is conserved. In other words, if all the forces pushing or pulling at an object cancel each other, the net force is zero. And, with a net force of zero, the object's motion is unchanged. If it was motionless, it remains motionless. If it was moving in a certain direction at a certain speed, these remain the same. (We say its velocity remains the same since velocity means a certain speed in a certain direction.) And, of course, if the object's mass and motion remain unchanged, the product of mv will remain unchanged - the momentum is unchanged. We would write the mv = 0 or the p = 0.

Impulse An object experiences an impulse when a force is applied to it for a certain length of time. In wrestling, the opponents push against each other for several seconds at a time. In baseball, the bat pushes against the ball for fractions of a seconds. The force causes a change in the object's motion and so its momentum changes. Think: you are walking along and someone pushes against your chest - the impulse you received from them slows you a bit and your momentum decreases. If they had pushed harder, you would have received a larger impulse, you would have slowed more and your momentum would have become even less. With a harder push still, the impulse would have stopped you. With a very hard push (with a very large impulse) you would have been stopped and caused to go backwards. So, the most extreme example of momentum change occurs when an object's direction is reversed by the impulse from a collision. We write I = Ft, where I is the impulse in Ns, F is the applied force and t is the length of time the force is applied. Recall that the I causes a change in the object's momentum because its velocity is changed and so we write I = p.


Q. What impulse results from a 10 N force applied to a 20 kg object for a time interval of 0.2 s?
I = Ft so I = 10 N * 0.2 s = 2 Ns.

Q. The object's momentum will change by how much as a result of the impulse?
I = p so the momentum change, p, = 2 kg*m/s.



Q. What happens to the object's velocity? It depends on the direction of the impulse; did it speed up or slow down the object. If the object moves faster after the impulse, its momentum will rise. If the object moves more slowly after the impulse, is momentum will have decreased.

Say the object was moving to the right at 5 m/s. Its p = mv = 20 kg * 5 m/s = 100 kg*m/s. If the impulse pushes the object forward, its new momentum is 100 kg*m/s + 2 kg*m/s or 102 kg*m/s. To find the new speed use p = mv. The v = p/m = 102 kg*m/s / 20 kg = 5.1 m/s.

If the impulse slowed the object, its new momentum is 100 kg*m/s - 2 kg*m/s = 98 kg*m/s. The new speed is v = p/m = 98 kg*m/s / 20 kg = 4.9 m/s.


Impulse-Momentum Theorem As we have just seen above, impulse and momentum changes are tied closely together. This fact is expressed in a formula called the Impulse-Momentum Theory. We write Ft = p, i.e., an impulse = a momentum change.


Applications of the Impulse-Momentum Theory

1. Why Does It Hurt? Two things can determine if an impact hurts us: how much our momentum changes and the time interval during which the change occurs. For example, a skier sliding quickly into a tree undergoes a large momentum change because the velocity change is large. They may be injured because the inflexible trunk causes the deceleration to occur in a fraction of a second. So what? Remember that a large p means the Ft will be large. Since the skier decelerates quickly, the t is small but since the Ft is a large value, there must be a large F of deceleration pushing against the skier. Because our body can withstand only small forces, pain and damage occur. If the same situation occurs with a safety net, the time interval is much longer, allowing for a smaller F and therefore no bodily injuries. Do the math!

Q. A 70 kg skier moving at 7 m/s hits a tree and, after just 0.9 s is stopped. The tree exerted what deceleration force against the skier's body?

Before collision: p = mv = 70 kg * 7 m/s = 490 kg*m/s

After collision: p = 70 kg * 0 m/s = 0 kg*m/s. See that the momentum has decreased by 490 kg*m/s

p = 490 kg*m/s and, since Ft = p, the Ft = 490 kg*m/s

F * 0.9 s = 490 kg*m/s so F = 490 kg*m/s / 0.9 s = 544.4 N pushing against the skier. No wonder the skier will be hurting.


2. Smokin'! Brake pads on large vehicles are not large just because the vehicle's wheels are larger but to create a larger breaking force. When a heavy truck stops, it undergoes a larger momentum change (a larger p and therefore a larger Ft) than a small, light car. But, if it is to stop in the same time interval as the car, a much larger deceleration force must be applied and this requires larger brake pads.

During successive brake performance tests, a vehicle may be brought to a stop from different speeds in the same time interval. Decelerating to zero from higher and higher speeds creates larger and larger changes in the vehicle's momentum. But, if the braking time, the t, is the same, ever larger deceleration forces from the brakes are required. So, if the p continues to increase but the t remains constant, the F (the deceleration F from the brakes) increases also. For this reason, high performance vehicles (which often have to stop quickly from excessive speeds) may also have large brake systems.

Angular Momentum This is the quantity of motion, equivalent to linear momentum, of an object moving in a circle. Use pANG = mvr.

Applications of Angular Momentum

1. I feel Dizzy! A skater can change their rotational speed just by holding their arms in or out. Theirlegmuscles can cretae Once into their rotation, the pANG remains constant. If the r is changed by altering arm extension, the v changes inversely. So, "arms in" causes rapid rotation, and visa versa.

2. My, That Season Flew By! Recall that Kepler's Laws state that a planet moves most quickly when it is closest to the sun. The same idea is revisited in terms of angular momentum. Like the skater, a planet has a constant pANG. During the part of its orbit around the sun when its r decreases, its v increases.

Conservation of Momentum When objects collide, speeds and directions can change, i.e., velocities can change. These changes occur because the collision causes impulses. In a two object collision, the momentum lost by one object equals the momentum increase of the other. Use pA + pB = pA' + pB' or mAvA + mBvB = mAvA' + mBvB' Sometimes, the colliding objects become locked together after the collision, like the cars of a train as it is being assembled in the rail yard. then, the objects will of course have the same post-collision speed. In this case, we write mAvA + mBvB = (mA + mB) v

Linear Collision Situations Here is a collection of situations in which objects bash together would experience velocity changes and therefore momentum changes. Can you think of others?

1. Catch up: a) an object in the rear moves faster than one in front and "rear ends" it. The rear object slows and the front object speeds up.
b) object in rear stops, object in front speeds up.

c) object in rear reverses its direction, the front object speeds up.



2. Head on: a) one object reverses its direction, the other slows.
b) both reverse direction.
c) one stops, the other reverses its direction.
d) they stick and move together, one reversing its direction.

Two Dimensional Collisions When objects collide, they usually do not stay in their vectors but may ricochet at some angle to the original paths. Imagine a cue ball moving E just before it collides with a second, motionless billiard ball. After the collision, one ball moves NE and the SE. Conveniently, a triangle can be used to visually describe the situation. Its base represents the pre-collision momentum (of the cue ball) and its two arms represent the post-collision momenta (of the two balls.) The left arm angles up to the apex to represent the final momentum of the ball moving NE and the right arm angles down from the apex to represent the final momentum of the ball moving SE. The triangle can exist because the original momentum equals the sum of the final momenta; when the second arm is actually attached to the end of the first, the space created for the triangle's base is exactly the length to represent the initial momentum of the cue ball. Because we know the cue ball's initial momentum, we know how long to make the base. Then, using the actual collision angles, trigonometry helps us find the lengths of the two arms, i.e., the final momenta of the billiard balls. Once we know the balls' final momenta, we can find their final velocities from p = mv. Recall that, during the collision, the cue ball loses energy to the second ball and so after the collision, the velocity of either ball is less than that of the cue ball before the collision.



Elastic or Inelastic Collisions In an elastic collision, the colliding objects do not lose momentum through deformation and friction. Their initial and final kinetic energies are constant. In inelastic collisions, KE values of the objects after the collisions are less than the KE values of the objects before the collisions because some of the objects' KE is "wasted" as heat or sound or inefficient deformations. We can write KE + KE = KE + KE