AP Momentum
and Impulse

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I. Defining Momentum

Review – Momentum is a mathematical concept useful in _____________________

Momentum is _________________ in all interactions

Momentum Basics

Momentum -

a) defined – symbol for momentum
b) Scalar? Or Vector? Direction? ___________________
c) $\Delta p = $
d) Force and momentum: Alter to $F_{net} = ma$ to find the relationship between $F$ and $\Delta p$

Impulse

e) Impulse = =

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

Teacher Key: I. Fundamentals

Missing Words: (Review) describing collisions and explosions | (Conservation) conserved | (a) product of mass and velocity, Symbol: $p$ | (b) Vector, direction of velocity | (c) $m \Delta v$ | (d) $F_{net} = \frac{\Delta p}{\Delta t}$ | (e) $J = \Delta p = F_{avg} \Delta t$.

Reasoning: Momentum change is caused by an impulse, which is graphically the area under a Force-Time plot.

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?

Teacher Key: Ex 1

Equations: $J = F_{avg} \Delta t = \Delta p$

Step-by-Step: Since $\Delta p$ (and thus final velocity) depends on the product of Force and Time, a smaller force can result in a larger velocity if the ball remains in contact with the strings for a much longer duration.

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

Teacher Key: Ex 2

Equations: $F_{net} = \frac{m \Delta v}{\Delta t}$

Step-by-Step: When a passenger stops, $m \Delta v$ is a fixed quantity. Airbags are designed to increase the time of the collision ($\Delta t$). Mathematically, since $\Delta t$ is in the denominator, increasing it significantly reduces the average net force ($F_{net}$) exerted on the passenger's body.

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 Key: Ex 3

Claim: (c) provides the largest transfer.

Step-by-Step: Momentum is a vector. To stop the ball (Case A), $\Delta p = m(0 - v) = -mv$. To reverse the ball (Case C), $\Delta p = m(-v - v) = -2mv$. Reversing direction requires twice the impulse, which the player must receive in return according to Newton's 3rd Law.

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 $7.0 \times 10^{-4} \, \text{s}$?

Teacher Key: Ex 4

Solution: $$F_{avg} \Delta t = m \Delta v$$ $$F_{avg} (7.0 \times 10^{-4} \, \text{s}) = (2.0 \, \text{kg})(0 \, \text{m/s} - 6.0 \, \text{m/s})$$ $$F_{avg} = \frac{-12 \, \text{kg} \cdot \text{m/s}}{7.0 \times 10^{-4} \, \text{s}}$$

Result: $1.71 \times 10^4 \, \text{N}$ (directed opposite to motion).

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 $2.0 \times 10^{-3} \, \text{s}$

Teacher Key: Ex 5

Solution: (a) $J = m \Delta v = (0.15 \, \text{kg})[22.0 \, \text{m/s} - (-20.0 \, \text{m/s})] = 6.30 \, \text{N} \cdot \text{s}$
(b) $F_{avg} = \frac{J}{\Delta t} = \frac{6.30 \, \text{N} \cdot \text{s}}{2.0 \times 10^{-3} \, \text{s}}$

Result: (a) $6.30 \, \text{N} \cdot \text{s}$ | (b) $3150 \, \text{N}$.

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)

Teacher Key: Ex 6

Phase 1: Energy Principles to find Impact Velocity $$U_{gi} = K_f \implies mgh = \frac{1}{2} m v^2$$ $$10 \, \text{m/s}^2 \cdot 3.00 \, \text{m} = \frac{1}{2} v^2 \implies v = \sqrt{60} = 7.75 \, \text{m/s}$$

Phase 2: Impulse-Momentum Theorem for Force $$F_{net} \Delta t = m(v_f - v_i)$$ $$F_{net} (0.550 \, \text{s}) = (82.0 \, \text{kg})(0 \, \text{m/s} - (-7.75 \, \text{m/s}))$$ $$F_{net} = \frac{635.5 \, \text{kg} \cdot \text{m/s}}{0.550 \, \text{s}}$$

Result: $1155.5 \, \text{N}$ Upward.

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 Key: Ex 7

Solution: $$a = \frac{F_{net}}{m} = \frac{8410 \, \text{N} \text{ (East)}}{2240 \, \text{kg}} = 3.75 \, \text{m/s}^2 \text{ (East)}$$ $$v_f^2 = v_i^2 + 2ad$$ $$(5.00 \, \text{m/s})^2 = (-20.0 \, \text{m/s})^2 + 2(3.75 \, \text{m/s}^2)d$$ $$25 = 400 + 7.5d \implies d = -50.0 \, \text{m}$$

Result: $50.0 \, \text{m}$ West.

IV. Conservation Principles

Watch Conservation Video (1-6)

When 2 objects interact, the ________________ momentum before the interaction equals __________________________________________

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

Teacher Key: Conservation Theory

Missing Words: (Blank 1) total | (Blank 2) the total momentum after | (Blank 3) Third.

Reasoning: Newton's 3rd Law ensures forces are equal and opposite, which means the impulses are equal and opposite, resulting in zero net change for the system momentum.

Derivation Workspace ($F_{net} = F_{net}$)

Case I – Inelastic Collision

Objects __________________. Assume no friction.

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

Teacher Key: Inelastic

Missing Words: stick together | not conserved | mechanical energy is converted to internal energy (heat/deformation).

Case II – Elastic Collision

Elastic Collision / ___________________________________
Review KE = _______

Teacher Key: Elastic

Missing Words: Kinetic energy is conserved | $\frac{1}{2} m v^2$.

Case III – PERFECT Elasticity

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?

Teacher Key: 1a

Solution: $$m_b v_i = (m_b + m_w) v_f$$ $$(0.0080 \, \text{kg}) v_i = (0.0080 + 9.0 \, \text{kg})(0.40 \, \text{m/s})$$

Claim: $450.4 \, \text{m/s}$.

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

Teacher Key: 1b

Solution: $$F_{avg} = \frac{m \Delta v}{\Delta t} = \frac{0.0080 \, \text{kg} (0.40 - 450.4 \, \text{m/s})}{0.0030 \, \text{s}}$$

Claim: $1201.1 \, \text{N}$.

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

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

Teacher Key: 1cd

Claim: (c) $1201.1 \, \text{N}$ | (d) $0 \, \text{kg} \cdot \text{m/s}$.

Reasoning: (c) Newton's 3rd Law. (d) Momentum is conserved for a closed 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 $7.0 \times 10^{-4} \, \text{s}$?

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?

Teacher Key: 3

Solution: $$m_1 v_1 + 0 = (m_1 + m_2) v'$$ $$(40000 \, \text{kg})(5.0 \, \text{m/s}) = (70000 \, \text{kg}) v'$$

Result: $2.86 \, \text{m/s}$.

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 impactsituations:
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?

Teacher Key: 5

Solution 5b: $$\Delta K = [\frac{1}{2}(200)(2.5)^2 + \frac{1}{2}(5000)(0.1)^2] - 0$$ $$\Delta K = 625 + 25 = 650 \, \text{J}$$

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.

VI. Recoil & 2D Analysis

Watch Recoil Video Watch More Conservation 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?

Teacher Key: Recoil 1a

Solution: $$\sum p_{ix} = \sum p_{fx}$$ $$0 = (875 \, \text{kg})v_c + (35.0 \, \text{kg})(145 \cos 35^\circ)$$

Claim: $4.75 \, \text{m/s}$ Recoil.

(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 $t = 2.0 \times 10^{-3} \, \text{s}$
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?