Momentum and Impulse Practice Problems

 

Momentum

Momentum can be defined as "mass in motion." All objects have mass; if an object is moving, then it has some momentum. The amount of momentum that an object has depends on upon two variables: how much mass the moving object has, and how fast the object is moving. In terms of an equation, the momentum of an object is equal to the mass of the object multiplied by the velocity of the object. Therefore, momentum is a vector and it points the same direction as the velocity. Momentum is always conserved (the total amount of momentum in the system stays the same over time) unless an external unbalanced force acts upon the system. Momentum is very abstract which is why we need to build mental models to understand it. 

 

Impulse

When determining momentum, you must first define your system. This is very similar to how it was important to define your system with energy too. Impulse is nothing but change in momentum (not to be confused with work, which is the change of energy in a system). Impulse is to momentum as work is to energy. Impulse is always caused by an unbalanced external force exerted on the system over time. Impulse is a vector quantity, so be sure to give direction in words. 

 

Impulse Example: Happy vs. Sad Ball Demo:

 

The happy ball and sad ball, both made of rubber, have the same mass. However, when they are dropped from the same height, the happy ball bounces back up and the sad ball sticks to the ground. The instant before they hit the ground, the balls have the same momentum since p=mv and their masses and velocities are the same (since all objects have the same acceleration in freefall and both balls start at a velocity of 0 m/s). Both the happy and sad balls have a negative momentum (going downwards) as they fall towards the ground. Then, upon hitting the ground, the happy ball bounces back up and since p=mv, the momentum must be in the same direction as the velocity. Therefore, the happy ball's final velocity is positive (as it is moving upwards). On the other hand, the sad ball ends with a velocity of 0 m/s because it sticks to the ground and as a result, it has a momentum of 0 kg*m/s. Impulse is equal to the change in momentum, or pf-pi. This demo stresses the importance of signs when it comes to momentum. As the analyzers, it is important we remember to recognize which direction the momentum going.

 

 

Common misconception

 

Explanation: 

Assume the balls have a mass, m, and their velocities the instant before hitting the ground are v.

 

Happy Ball:

Sad Ball:

 

Therefore, the impulse of the happy ball is greater than the impulse of the sad ball.

 

 

Momentum/Impulse Equations

 

Variable for Momentum is "p", do not confuse it with "m", which is the variable for mass.

 

Momentum:                    p = mv

 

Impulse:                          △p = P_{f(Final Momentum)} - P_{i(Initial Momentum)} = ΣF_ext△t

 

 

Momentum does not have a single letter abbreviation for it's units. Instead it can be written in two units, both of which are equal:

     (N) * s [newton seconds]

     (kg) * (m/s) [kilogram meter per second]

 

We know from Newton's 2^nd Law: ΣF = ma. We also know a = △v/△t. By substitution then:

 

ΣF = m△v / △t           and since p = mv

 

Therefore:

 

Conservation of Momentum

One of the most important ideas in physics is the law of conservation of momentum. In words, as long as no external unbalanced forces act on a system, the total momentum remains the same. Do not just blindly assume that momentum will be conserved. First a draw a free body diagram for your system. Then determine if your forces are internal or external forces. Any force acting on the system by something else inside the system is an internal force and will not move move momentum in/out of a system (since both force pairs will be in the system & they will end up canceling each other out). However if a force acts on the system by an object outside the system then that's an external force (since only one of the force pairs will be in the system). If there is only one external force, or the external forces are unbalanced the total momentum of the system will change. However if the external forces sum to zero then momentum will be conserved. Another way to think about this rule is for a system of the entire universe. Since there cannot be an external force on the whole universe, momentum in the entire universe is always conserved (i.e. the total stays the same). 

 

Mathematically, the law of conservation of momentum can be stated: 

 

If            then           

 

 

Example 1:

Suppose two people on skateboards are standing motionless on a smooth surface 2.0 m apart. Person B throws a 5.0 kg ball to person A. The ball is traveling at a speed of 10. m/s. How fast does person B move after throwing the ball? 

 

Part II:

How fast does person A go after catching the ball?

 

Example 2: 

Consider a situation where a 15 kg ball is thrown at a velocity of 20 m/s to a 60-kg person who is at rest on ice. The person catches the ball and subsequently slides with the ball across the ice. Determine the velocity of the person and the ball after the collision.

 

 

 

Such a motion can be considered as a collision between a person and the ball. Before the collision, the ball has momentum and the person does not. The collision causes the ball to lose momentum and the person to gain momentum. After the collision, the ball and the person travel with the same velocity across the ice.

 

Creating a table (alternate method to solve): 

A new way to solve a problem like this is to make a table. Start off by creating two columns that represent each major stage of the problem. In this case, we are only focusing on the before and after of the collision. Then add a row for each thing in the system. In this case, we should label the person and the ball. Finally, add a row for the total. 

 

Solution:

First off, state your system as the person, the ball, Earth, and the ice.

 

Now, for each table cell, remember to write down the equation in use. Since we are focusing on momentum, p=mv will be the equation we use in the entire problem.

 

For the person before the collision, we can substitute their mass with 60kg and the velocity with 0m/s. 60kg * 0 m/s = 60kg m/s.

 

Next, move on to the ball before the collision. You can now substitute the ball's mass with 15 kg and the ball's velocity with 20 m/s. 15kg * 20 m/s = 300 kg m/s.

 

We can now fill in the total before the collision. We can just add the momentum's of each object, and in this case, we get 0 kg m/s + 300 kg m/s = 300 kg m/s. This means that before the collision occurs, the net momentum of the system is 300 kg m/s right. 

 

Now, we can move on to the "After Collision" column. For the person, we can substitute the mass with 60kg. Since, the final velocity for the person is not given, we can leave it as a variable. So the final momentum of the person is 60kg * Vf_man  

 

Next, move on to the ball after the collision. Once again, substitute 15 kg for the ball's mass, and since we don't know the final velocity of the ball, we can leave it as V_fball

So the final momentum for the ball is 15 kg * V_fball.

 

Finally, the total momentum after the collision must also be 300 kg m/s because momentum is conserved. We know this because there aren't any net unbalanced forces acting on the system.

 

 

Now, let's solve for the final velocities. 

 

First off, we know that the final velocities of the ball and the person are the same. 

 

We can show the net final velocity algebraically by adding the final momentum of the ball and the person. The unit of mass for both the person and the ball are the same, so we can add the masses 60kg and 15kg to get 75kg. 

 

We can also write their net velocity as one variable (v). 

 

Our final equation can be modeled as 75kg (v) = 300 kg m/s. By dividing both sides by 75kg, we can isolate the velocity, and you should get 4 m/s.

 

 

Force vs. Time Graphs

The area inside the line of a force vs. time graph is the impulse/change in momentum. Since the force is usually on the y axis and the time is usually on the x axis, you can calculate the impulse based on this graph by finding the area of the shaded region (the external forces multiplied by the time). 

 

 

Problem Solving Approach for Momentum/Impulse Questions

 

Step 1: Define the system.

 

Step 2: Define negative and positive direction (important since momentum is a vector!)

 

Step 3: Determine if momentum is conserved.

If ∑F_ext = 0, momentum is conserved as there are no external forces on the object.

 

Step 4: Write a summation of momentum equations at critical points in time (i.e. before/after the collision). It's easy to write a summation of momentum equation. Simply put a momentum term for each object in your system. Your summation of momentum equations should look like:

∑p_i = p_a + p_b + p_c + etc.    (every object in the system)

∑pf = p_a + p_b + p_c + etc.    (every object in the system)

 

Step 5: Find the total momentum (∑P) at some point in time. This is valuable since if momentum is conserved, once you know the total momentum anywhere you know the total momentum everywhere. Even if the total momentum is not conserved, knowing the total momentum at one point is useful for finding the momentum somewhere else as long as you know how the momentum of the system changed (i.e. what the impulse was). 

 

Step 6: Use your summation of momentum equations to find any unknown quantities. Remember that if two objects stick together they must have the same velocity and their momentum term can be combined into something like:

∑p_i = p_a+b+ p_c + etc.    (every object in the system)

 

Step 7: Make sure to write direction on all vectors like momentum and impulse. Then check if your answer is reasonable or not. 

 

 

Screencast: Problem Solving Momentum using the 5 step method 

 

 

Comparing and Contrasting Energy and Momentum

 

Energy and momentum are very analogous in many ways; however, they also have a few major differences. 

 

 

 

 

Elastic & Inelastic Collisions

 

The difference between elastic and inelastic collisions is: if two or more objects collide and they stick together, it is inelastic, if they do not stick together, the collision is elastic.

 

Momentum Kahoot

 

https://create.kahoot.it/share/momentum-quiz/d8237afb-11ca-4aff-bb96-91528376b169

 

 

Momentum Graph Builder