Unit 4: Linear Momentum

Comprehensive Review & Exam-Style Practice

Section 4.1

Linear Momentum

Momentum is a vector quantity defined as the product of mass and velocity: \( \vec{p} = m\vec{v} \). Newton’s Second Law can be re-defined as the time rate of change of momentum: \( \vec{F} = \displaystyle\frac{d\vec{p}}{dt} \).

Q1: A particle of mass \( m \) has momentum of magnitude \( p \). Which of the following represents its kinetic energy?

(A) \( p^2/m \)
(B) \( p^2/2m \)
(C) \( p/2m \)
(D) \( pm \)

Kinetic energy is \( K = \displaystyle\frac{1}{2}mv^2 \).
Substitute \( v = p/m \) into the energy equation:
\( K = \displaystyle\frac{1}{2}m\left(\displaystyle\frac{p}{m}\right)^2 = \displaystyle\frac{1}{2}m\left(\displaystyle\frac{p^2}{m^2}\right) = \displaystyle\frac{p^2}{2m} \).

Correct Answer: B

Q2: If the momentum of a particle is given by \( p(t) = 4t^3 – 2t \), what is the magnitude of the net force acting on it at \( t = 2 \text{ s} \)?

(A) 14 N
(B) 28 N
(C) 46 N
(D) 50 N

Net force is the first derivative of momentum:
\( F = \displaystyle\frac{dp}{dt} = 12t^2 – 2 \).
Evaluate at \( t = 2 \):
\( F(2) = 12(2^2) – 2 = 48 – 2 = 46 \text{ N} \).

Correct Answer: C

Section 4.2

Impulse and Change in Momentum

Impulse (\( \vec{J} \)) is defined as the change in momentum: \( \vec{J} = \Delta \vec{p} = \int \vec{F} dt \). For a constant average force, \( \vec{J} = \vec{F}_{avg} \Delta t \).

Q1: A 0.5 kg ball hits a wall at 10 m/s and bounces back at 8 m/s in the opposite direction. What is the magnitude of the impulse?

(A) 1 Ns
(B) 4 Ns
(C) 5 Ns
(D) 9 Ns

Let the initial direction be positive.
\( v_i = 10 \text{ m/s} \), \( v_f = -8 \text{ m/s} \).
\( J = \Delta p = m(v_f – v_i) = 0.5(-8 – 10) = -9 \text{ Ns} \).
The magnitude is 9 Ns.

Correct Answer: D

Q2: A force \( F(t) = At^2 \) acts on a mass \( m \) for a duration \( T \). What is the total impulse?

(A) \( AT^2 \)
(B) \( 2AT \)
(C) \( \displaystyle\frac{1}{3}AT^3 \)
(D) \( AT^3 \)

Impulse is the integral of force over time:
\( J = \int_0^T At^2 dt = [\displaystyle\frac{1}{3}At^3]_0^T = \displaystyle\frac{1}{3}AT^3 \).

Correct Answer: C

Q3: An object’s momentum increases linearly from 5 kg·m/s to 15 kg·m/s over 2 seconds. What is the average net force?

(A) 5 N
(B) 10 N
(C) 20 N
(D) 30 N

\( F_{avg} = \displaystyle\frac{\Delta p}{\Delta t} = \displaystyle\frac{15 – 5}{2} = \displaystyle\frac{10}{2} = 5 \text{ N} \).

Correct Answer: A

Section 4.3

Conservation of Linear Momentum

In an isolated system where the net external force is zero, the total momentum remains constant: \( \sum \vec{p}_i = \sum \vec{p}_f \).

Q1: A 1000 kg cannon at rest fires a 10 kg ball at 200 m/s. What is the recoil velocity of the cannon?

(A) -0.2 m/s
(B) -2 m/s
(C) 0.2 m/s
(D) 2 m/s

Initial momentum is zero. Final momentum must be zero:
\( 0 = (10)(200) + (1000)v_{cannon} \).
\( -2000 = 1000v_{cannon} \Rightarrow v_{cannon} = -2 \text{ m/s} \).

Correct Answer: B

Q2: A 5 kg object at rest explodes into two pieces. A 2 kg piece moves North at 10 m/s. What is the velocity of the 3 kg piece?

(A) 6.67 m/s South
(B) 6.67 m/s North
(C) 15 m/s South
(D) 15 m/s North

Momentum is conserved: \( P_i = 0 \). Let North be positive.
\( 0 = (2)(10) + (3)v_2 \).
\( -20 = 3v_2 \Rightarrow v_2 = -6.67 \text{ m/s} \).
The negative sign denotes the South direction.

Correct Answer: A

Q3: If two objects collide and stick together in an isolated system, which of the following remains constant?

(A) Total kinetic energy
(B) Velocity of each object
(C) Velocity of the center of mass
(D) Total internal energy

In an isolated system, the acceleration of the center of mass is zero because there are no external forces (\( \sum F_{ext} = M\vec{a}_{cm} = 0 \)). Thus, \( \vec{v}_{cm} \) must remain constant.

Correct Answer: C

Section 4.4

Elastic and Inelastic Collisions

In all isolated collisions, momentum is conserved. Elastic collisions also conserve kinetic energy (\( K_i = K_f \)), while inelastic collisions lose kinetic energy to internal forms (heat, sound, etc.).

Q1: A 2 kg block moving at 6 m/s sticks to a 4 kg block at rest. What is the loss in kinetic energy?

(A) 12 J
(B) 24 J
(C) 36 J
(D) 48 J

Initial KE: \( K_i = \displaystyle\frac{1}{2}(2)(6^2) = 36 \text{ J} \).
Momentum conservation: \( (2)(6) = (2+4)v_f \Rightarrow v_f = 2 \text{ m/s} \).
Final KE: \( K_f = \displaystyle\frac{1}{2}(6)(2^2) = 12 \text{ J} \).
Loss: \( 36 – 12 = 24 \text{ J} \).

Correct Answer: B

Q2: A particle of mass \( m \) moving with speed \( v \) undergoes a perfectly elastic head-on collision with an identical particle at rest. What are the final speeds?

(A) Both move at \( v/2 \)
(B) The first stops, the second moves at \( v \)
(C) The first moves at \( v \), the second remains at rest
(D) Both move at \( v/\sqrt{2} \)

In a 1D elastic collision between equal masses where one is at rest, the moving mass transfers all its momentum and energy to the stationary one. They essentially “swap” velocities.

Correct Answer: B

Q3: In a collision where the objects bounce off each other but kinetic energy is not conserved, which of the following is true?

(A) Momentum is conserved, but kinetic energy is lost.
(B) Momentum is lost, but kinetic energy is conserved.
(C) Both momentum and kinetic energy are conserved.
(D) Both momentum and kinetic energy are lost.

In an isolated system, momentum is always conserved regardless of the collision type. “Inelastic” simply means that kinetic energy is not conserved (it is converted to other forms).

Correct Answer: A

Momentum Formulas Recap

Momentum\( \vec{p} = m\vec{v} \)

Newton II\( \vec{F} = \displaystyle\frac{d\vec{p}}{dt} \)

Impulse\( \vec{J} = \Delta \vec{p} = \int \vec{F} dt \)

KE relationship\( K = \displaystyle\frac{p^2}{2m} \)

Conservation\( \sum \vec{p}_i = \sum \vec{p}_f \)

CM Velocity\( \vec{v}_{cm} = \displaystyle\frac{\sum m_i \vec{v}_i}{M_{total}} \)

AP Physics C Mechanics Quick Review: Unit 4 Linear Momentum