AP Momentum
and Impulse

Momentum Header Graphic

showmethephysics.com

I. Defining Momentum

Momentum / Impulse

Review – Momentum is a mathematical concept useful in _____________________

Momentum is _________________ in all interactions

Momentum Basics

Momentum -

Impulse

e) Impulse _____ = _______ = __________

Find the $\Delta p$ for an object with the F vs t plot:

Force vs Time Graph
Teacher CER Key: Module I
Evidence (Substitution) Equation: $F_{net} = ma = m\frac{\Delta v}{\Delta t} = \frac{\Delta p}{\Delta t} \implies J = F \Delta t = \Delta p$. Units: $kg \cdot m/s$ or $N \cdot s$.
Reasoning Momentum describes an object's "quantity of motion." Impulse is the change in that momentum caused by a force applied over a specific time interval. Graphical area under $F$ vs $t$ equals $\Delta p$.
Claim (Review) describing collisions/explosions. (Conservation) conserved. (a) product of mass and velocity; symbol: $p$. (b) Vector; direction of velocity. (c) $m\Delta v$. (e) Impulse $J = \Delta p = F_{avg} \Delta t$.
↑ Back to Main Menu

II. Conceptual Analysis

Ex 1) Two tennis players hit an overhead smash on the ‘sweet spot’ of their tennis rackets. The player who returned the ball with the greatest velocity applied a smaller force than the other player. How is this possible?

Ex 2) Explain why automobiles have airbags, and describe how they work. Include the relevant equations.

Ex 3) A hockey player is standing still on a frictionless frozen lake. Her friend throws a bowling ball straight at her. In which of the following cases is the largest momentum transferred to the hockey player?
(a) The hockey player catches the bowling ball and holds on to it.
(b) The hockey player catches the bowling ball momentarily, but then drops it vertically downwards.
(c) The hockey player catches the bowling ball, holds it momentarily, and throws it back to her friend.

Teacher CER Key: Module II
Evidence Ex 1: $J = F\Delta t$. Ex 2: $F = \frac{\Delta p}{\Delta t}$. Ex 3: $\Delta p = m(v_f - v_i)$.
Reasoning Ex 1: A smaller force applied over a much longer contact time can result in a larger impulse. Ex 2: Airbags increase $\Delta t$, which decreases $F$ for a fixed $\Delta p$ (stopping the passenger). Ex 3: Bouncing an object back requires a larger $\Delta v$ (e.g., $v$ to $-v = 2v$ change) compared to just stopping it.
Claim Ex 1: Increase in contact time. Ex 2: Increased stopping time reduces force. Ex 3: (c) Throwing it back.
↑ Back to Main Menu

III. Practice Calculations

Ex 4) A 2.0-kilogram lead brick is moving at a velocity of 6.0 m/s. What size force is needed to bring the lead brick to rest in $7.0 \times 10^{-4}$ seconds?

Ex 5) A 0.15 kg ball is thrown with a speed of 20.0 m/s. It is hit straight back at the pitcher with a final speed of 22.0 m/s.
(a) What is the impulse delivered by the bat to the ball?
(b) Find the average force exerted by the bat on the ball if the two are in contact for $2.0 \times 10^{-3}$ s

Ex 6) An 82.0-kilogram Olympic diver drops from rest from a diving board 3.00 m above the surface of the water and comes to rest .550 s after reaching the water. What is the net force acting on the diver as he is brought to rest? (Magnitude and direction)

Ex 7) A 2240 kg car traveling west slows down uniformly from 20.0 m/s to 5.00 m/s. If the force acting on the car was a constant 8410 N to the east, how far and in what direction does the car move during the deceleration?

Teacher CER Key: Module III
Evidence (Substitutions) Ex 4: $F = \frac{(2.0\,kg)(0\,m/s - 6.0\,m/s)}{7.0 \times 10^{-4}\,s} = -17,143\,N$.
Ex 5a: $J = (0.15\,kg)(22.0\,m/s - (-20.0\,m/s)) = 6.3\,kg \cdot m/s$.
Ex 5b: $F = \frac{6.3\,kg \cdot m/s}{2.0 \times 10^{-3}\,s} = 3150\,N$.
Ex 6: $v = \sqrt{2(9.8\,m/s^2)(3.0\,m)} = 7.67\,m/s$. $F = \frac{(82.0\,kg)(0 - (-7.67\,m/s))}{0.550\,s} = 1143\,N$.
Reasoning Net force is the rate of change of momentum. By knowing mass, change in velocity, and time, we can calculate average force required for the interaction.
Claim Ex 4: $1.71 \times 10^4\,N$. Ex 5: (a) $6.3\,N \cdot s$ (b) $3150\,N$. Ex 6: $1143\,N$ Upward. Ex 7: $50.0\,m$ West.
↑ Back to Main Menu

IV. Conservation Principles

Watch Conservation Video (1-6)

Conservation of Momentum - When 2 objects interact, the _____________ momentum before the interaction equals __________________________________________

The conservation of momentum comes from Newton’s ____________ Law.

Derive the conservation of momentum from:
$F_{net} = F_{net}$

Finding Total Momentum / Assume no friction

Finding Total Momentum Diagram

Case I – Inelastic Collision

Objects __________________. Assume no friction.

Inelastic Collision - Stick Together

Kinetic Energy in a Collision: In an inelastic collision, KE is _________________________ because ________

Case II – Elastic Collision

Elastic Collision / ___________________________________
Review KE = _______

Elastic Collision - Bouncy

Case III – PERFECT Elasticity

KE not conserved in all problems – Which collision, if any, is perfectly elastic?

Case III Perfectly Elastic Diagram
Teacher CER Key: Module IV
Evidence Newton's 3rd: $F_{12} = -F_{21} \implies \Delta p_1 = -\Delta p_2 \implies \Delta p_{total} = 0$. Inelastic: $m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_{final}$.
Reasoning In a closed system, internal forces cancel. Momentum is always conserved. Kinetic Energy is only conserved in elastic collisions; it's lost to heat/sound/deformation in inelastic ones.
Claim (Blank 1) Total (Blank 2) total momentum after. (Law) 3rd. (Case I) Stick together. (KE Inelastic) Lost/not conserved. (Case II) Kinetic Energy is conserved. (Case III) Atoms/Molecules collisions.
↑ Back to Main Menu

V. Impulse/Collision Worksheet

Watch Inelastic Solutions

1. A bullet with a mass of .0080-kg (8.0-gram) is fired horizontally into a stationary 9.0-kg block of wood on a frictionless horizontal surface. The bullet sticks into the block of wood and after the impact, the block has a speed of 0.40 m/s.

a. What was the initial velocity of the bullet?

b. If the bullet came to rest inside the block in 3.0 milliseconds ($3.0 \times 10^{-3}$ sec), what force did the block of wood apply on the bullet?

100. What is the force on the block of wood?

d. What is the change in momentum of the bullet-block system?

2. A 2.0-kilogram brick is moving at a speed of 6.0 meters per second. How large a force is needed to bring the brick to rest in $7.0 \times 10^{-4}$ seconds?

3. A 40,000. kg freight car is coasting at a speed of 5.0 m/s along a straight track when it strikes a 30,000. kg stationary freight car and couples to it. What will be the speed after impact?

4. Two balls of equal mass, moving with a speed of 3.0 m/s, collide head-on. Find the velocity of each after the impact for each of the following situations.

    a. They have an inelastic (stick) collision.
    b. They have a perfectly elastic collision (KE conserved).

5. Four 50. kg girls are sitting in a boat at rest. They simultaneously dive horizontally in the same direction at 2.5 m/s from the same side of the boat. The empty boat has a speed of 0.1 m/s afterward.

    a. What is the mass of the boat?

    b. What is the change in mechanical energy of the system?

6. A 4,000. kg truck is moving eastward through a frictionless intersection at 3.00 m/s. A second truck, with a mass of 8,000. kg is moving northward through the same intersection at 2.00 m/s. They collide and stick together.

    a. What is their speed immediately after impact?

    b. What is the change in mechanical energy of the system?

    c. Name at least two places the energy went.

Teacher CER Key: Module V
Evidence (Substitutions) 1a: $(0.0080\,kg)v_i = (0.0080+9.0\,kg)(0.40\,m/s) \implies v_i = 450.4\,m/s$.
1b: $F = \frac{0.0080\,kg(0.40 - 450.4\,m/s)}{0.003\,s} = -1200\,N$.
3: $(40,000\,kg)(5.0\,m/s) = (70,000\,kg)v' \implies v' = 2.86\,m/s$.
Claim 1a: $450.4\,m/s$. 1b: $1200\,N$. 100: $1200\,N$. 1d: $0$. 3: $2.86\,m/s$. 4a: Both stop ($0\,m/s$). 4b: $v_1 = -3.0\,m/s$, $v_2 = 3.0\,m/s$.
↑ Back to Main Menu

VI. Recoil & 2D Analysis

Watch Recoil Video Watch Instruction Video

1. One of those Civil War cannons is fired. The cannon has a mass of 875 kg. It fires a 35.0 kg cannonball at a velocity of 145 m/s at an elevation angle of 35.0°. The length of the barrel is 2.10 m.

(a) What is the recoil velocity of the cannon?

(b) What is the KE of the cannonball as it leaves the cannon?

(c) How far does the cannonball travel in the horizontal direction?

(d) What is its momentum at the top of the cannon ball’s trajectory?

2. A 0.15 kg ball is thrown with a speed of 20.0 m/s. It is hit straight back at the pitcher with a final speed of 22.0 m/s.
(a) What is the impulse delivered by the bat to the ball?
(b) Find the average force if $t = 2.0 \times 10^{-3}$ s
c) Force exerted by the ball on the bat?

3. A railroad car of mass $2.00 \times 10^4$ kg moving at 3.00 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s.
(a) What is the speed of the three coupled cars after the collision?

Collisions in 2 Dimensions Worksheet

Watch 2D Instruction Page 1

Momentum is conserved in ________ and _______ directions

Total Mom in ___ direction before = Total Mom in ___ direction after

Total Mom in ___ direction before = Total Mom in ___ direction after

1. If the angle between 2 objects = 90 degrees ...
Add momentum vectors using .... ____________________________

1. Textbook - p. 190) 41

Collision in 2D - Textbook Problem 41 Vector Diagram for Problem 41

2. # 42 - Billiard Ball Instructional Video

Billiard Ball Table View Billiard Ball Vector Solution

Before $M_a$ =      $V_a$ =      $M_b$ =      $V_b$ =

After $M_a$ =      $V_a$ =      $\theta_a$ =      $M_b$ =      $V_b$ =

Total momentum X before = Total mom. x After?

Total momentum Y before = Total mom. Y after?

Teacher CER Key: Module VI
Evidence (Substitutions) Recoil 1a: $0 = (875\,kg)v_{rec} + (35.0\,kg)(145\,m/s \cdot \cos 35^\circ) \implies v_{rec} = -4.75\,m/s$.
1b: $K = \frac{1}{2}(35.0\,kg)(145\,m/s)^2 = 367,938\,J$.
1d: $p_{top} = (35.0\,kg)(145\,m/s \cdot \cos 35^\circ) = 4157\,kg \cdot m/s$.
Claim 1a: $4.75\,m/s$ West. 1b: $3.68 \times 10^5\,J$. 1d: $4.16 \times 10^3\,kg \cdot m/s$.
2D Theory: Horizontal and Vertical directions. X, X, Y, Y. Pythagorean Theorem.
↑ Back to Main Menu
showmethephysics.com - AP Momentum & Impulse Module