1
Q
What is classical mechanics?
A
The branch of physics that studies the motion of bodies.
This includes how objects move (kinematics) and why they move (dynamics). To explain why objects move, we need forces (interactions). ‘Classical’ means that we exclude quantum and relativistic effects.
This applies when objects are much larger than atoms and move much slower than the speed of light.
2
Q
List any four key physical quantities in classical mechanics.
A
- Force
- Acceleration
- Mass
- Energy
These are central to Newtonian mechanics and motion analysis. Others include position, velocity, linear momentum, work, angular velocity, angular acceleration, torque, and angular momentum.
3
Q
Define:
kinematics
A
The study of motion without considering its causes.
It essentially uses the tools of calculus (differentiation and integration) and, if needed, vectors. The objective is to link the position of an object to its velocity and acceleration.
Focuses on displacement, velocity, acceleration, and time.
4
Q
What are the main limitations of classical mechanics?
A
It fails if at least one of the following is true regarding the moving objects being studied:
- The object is moving very fast (its velocity is greater than around one-tenth the speed of light).
- The object is very small (around one nanometer or smaller).
- The object experiences very strong gravitational fields (such as near a black hole).
Quantum mechanics and relativity are needed for atomic systems, near-light speeds, or massive bodies like black holes.
5
Q
State the difference between instantaneous velocity and average velocity.
A
- Average velocity of an object over a time interval is the displacement of the object over the time interval divided by the time interval.
- Instantaneous velocity is the velocity at a specific point in time.
- As we look at the average velocity over smaller and smaller time intervals, we get the instantaneous velocity.
6
Q
Fill in the blank:
In uniformly accelerated motion, a velocity vs. time graph shows a straight line with a constant _____.
A
slope
The slope in a velocity-time graph represents the acceleration.
7
Q
What does the area under a velocity vs. time graph represent?
A
The displacement of the object.
This area gives the change in position over a time interval.
8
Q
True or False:
A curved position vs. time graph implies changing velocity.
A
True
Curvature in the graph indicates changing slope. Since the slope for the position-time graph is velocity, a changing slope shows changing velocity.
9
Q
Why is the acceleration vs. time graph a horizontal line for uniformly accelerated motion?
A
Because acceleration is constant over time.
A flat line indicates no change in acceleration (a = constant).
10
Q
The position of a particle moving horizontally is given by x(t) = 2.5t - 1.25t² (in meters) for t >= 0.
How would the position-time graph describe the motion of the particle?
A
Assuming that the axis points to the right, the particle initially moves to the right. It is initially moving at a velocity of 2.5 m/s, but it is slowing down.
It comes instantaneously to a stop at t = 1 second, and then starts moving to the left with increasing speed. It comes back to the origin at t = 2 seconds, and thereafter continues to move to the left with ever increasing speed.
11
Q
A particle moves with acceleration a(t)=kt, where k is a constant. Given the initial velocity (v₀) and the initial position (x₀ ) at the initial time (t = 0), find position as a function of time.
Assume constant acceleration.
A
Velocity is the integral of acceleration and then position is the integral of velocity. Make sure to take care of the limits of the integrations!
12
Q
A car moving along a straight road slows down uniformly from velocity v₀ to rest in time t. Which expression gives the acceleration?
Assume constant acceleration.
A
Since the acceleration is uniform, the acceleration is simply the final velocity minus the initial velocity followed by dividing by the time interval. In graphical terms, this is also the slope of the velocity-time graph (which is a straight line in this case with negative slope).
13
Q
The velocity of a falling object is given by the following equation: v(t)=gt
What is the position as a function of time?
g is the usual acceleration of free fall.
A
x₀ is the initial position.
This assumes constant acceleration due to gravity (no air resistance).
14
Q
An elevator moves from the bottom of an elevator shaft to the very top at a constant velocity of 1 m/s, covering a total distance of 300 meters.
At the instant that the elevator starts to go up, a coin is released (from rest) from the top of the shaft.
How long does the coin take to hit the elevator?
Ignore air resistance.
A
~7.65 seconds
The orange curve shows the coin; the blue line is for the elevator.
Note that once the coin hits the elevator, both move together. Here we are ignoring the size of the elevator.
To find the time taken, we just look for the intersection of the quadratic curve with the straight line. The former is given by H - 0.5gt². The latter is vt, where v is the elevator speed. Solving for t we get t = 7.65 seconds.
15
Q
A projectile lands at the same height it was launched from. What is the expression for the total flight time in terms of (v₀) and θ?
A
The vertical velocity just after launch is v₀ sin(θ). Ignoring air resistance, the vertical velocity just before impact on the ground is -v₀ sin(θ). This gives us the time of flight as T = (2v₀ sin(θ))/g.
16
Q
A projectile is launched with speed ( v₀ ) at angle θ above the horizontal. What is the expression for the maximum height it reaches?
A
The horizontal motion and vertical motion of a projectile are independent; for this problem, using the vertical motion only is sufficient. At the maximum height, the vertical velocity is zero, while the vertical velocity at the point of release is v₀ sin θ. The vertical acceleration is -g. Then, 0² = v₀² sin²θ - 2gH. Solve for H.
17
Q
Fill in the blank:
The horizontal range of a projectile launched on flat, horizontal ground with speed (v₀) and angle θ is given by ______.
A
Looking at the horizontal motion, the distance traveled is simply R = (v₀ cos(θ)) T where T = (2v₀ sin(θ))/g is the time of flight. The important point is that the horizontal velocity does not change (no horizontal acceleration since there is no horizontal force). Note that this answer would be different if the projectile was shot on a hill!
18
Q
True or False:
A projectile reaches its maximum horizontal velocity at the peak of its trajectory.
A
False
Horizontal velocity is constant in projectile motion (no horizontal acceleration since there is no horizontal force), so it does not vary.
19
Q
Two projectiles are launched on flat, horizontal ground with the same initial speed but at complementary angles (e.g., 30° and 60°).
How do their ranges compare?
Complementary angles are angles such that their sum is 90°.
A
They have the same range.
Let us call the two different launch angles α and β. Now, since they are complementary, α + β = 90°. The range with α is R = (v₀² sin(2α))/g. But sin(2α) = sin(2(90° - β)) = sin(180° - 2β) = sin(2β). Therefore, they have the same range.
20
Q
The velocity vector for an object is given by v = bt x̂ + ct² ŷ.
What is the acceleration vector?
Assume constant acceleration.
x̂ and ŷ are unit vectors, b and c are constants.
A
The acceleration vector is the derivative of the velocity vector. To differentiate a vector expressed using a Cartesian coordinate system, simply differentiate each of the components one by one.
21
Q
The velocity vector for an object is given by v = at x̂ + bt² ŷ.
What is the position vector?
x̂ and ŷ are unit vectors, b and c are constants.
A
The position vector is the integral of the velocity vector. To integrate a vector expressed using a Cartesian coordinate system, simply integrate each of the components one by one. Take care of the initial conditions (that is, the lower limits of the integrals) - that’s where the initial position vector r₀ comes in.
22
Q
How would you find the velocity vector for a projectile shot from flat, horizontal ground at an angle θ with respect to the horizontal?
A
One way to do this would be to treat the horizontal and vertical velocities independently using simple kinematics, and then just putting them together to write the vector. Alternatively, start from the acceleration vector a = -gŷ and integrate this (taking the initial velocity into account).
23
Q
What are Newton’s three laws of motion?
A
- First law: An object remains at rest or moves at constant velocity unless acted upon by a net force.
- Second law: F=ma, where Force equals mass times acceleration.
- Third law: For every action, there is an equal and opposite reaction.
Newton’s laws explain how forces affect motion; they form the foundation of classical mechanics.
24
Q
What are inertial frames?
A
Frames of reference where Newton’s first law holds true.
For example, in an accelerating train, a box on the floor appears to slide backward, even though no force is acting on it. This happens because the train is a non-inertial frame, where Newton’s laws don’t directly apply. To an observer on the ground (an inertial frame), the box simply continues moving at constant velocity while the
Newton’s first law is really important because it allows us to talk about inertial frames (and hence frames in which F = ma can be applied). It is not simply a special case of F = ma with F= 0.
25
What is the primary difference between gravity and most other forces in classical mechanics?
Most forces in classical mechanics arise from **contact between objects**, like friction, tension, and normal force.
The **main exception is gravity**, which acts at a distance without physical contact.
## Footnote
Electromagnetic forces are also such action-at-a-distance forces, but often electromagnetic forces cancel each other out since they can be both attractive and repulsive.
However, if two objects touch each other, these electromagnetic forces need not cancel each other out. As such, friction, tension, normal force, etc. are all fundamentally electromagnetic forces.
26
Explain how **mass and weight differ**, and their relationship to Newton's second law.
* **Mass** is a measure of an object's inertia; weight is the gravitational force acting on it.
* **Weight** is given by W = mg where g is the acceleration due to gravity.
## Footnote
Mass and weight have different units because they are different physical quantities. The former is measured in kg, while the latter in Newtons (in SI units).
Mass is constant, but weight can vary based on the local gravitational field.
27
# Define:
Net force
Vector sum of **all** forces acting on an object.
## Footnote
It determines the acceleration of the object according to F = ma where F is the net force.
Only the net force affects the motion of the object; individual forces may cancel each other out.
28
# Fill in the blank:
If an **object is accelerating**, there is a net force acting in the direction of \_\_\_\_\_\_\_.
acceleration
## Footnote
Newton’s Second Law defines this relationship. It is a vector equation, with the mass being a scalar. As such, the net force **F** has to be pointing in the same direction as the acceleration **a**.
29
What do **Newton’s laws** imply about the net force acting on an object moving at **constant velocity**?
The net force is **zero**.
## Footnote
Constant velocity (including rest) means zero acceleration, hence no net force.
30
Two objects experience the same net force. One has **twice the mass** of the other. How do the magnitudes of their **accelerations** compare?
The heavier object accelerates **half** as much.
## Footnote
From F=ma, acceleration is inversely proportional to mass if force is constant.
31
Explain the concept of a **free body diagram** and its utility in solving mechanics problems.
* It is a graphical representation **showing all the forces acting on a single body**.
* It **isolates the body** and helps in applying Newton's laws to determine **unknown forces or accelerations**.
## Footnote
What goes into Newton's second law are the forces acting **on** an object, not the forces exerted **by** the object. Free-body diagrams help to distinguish between these.
This tool is vital for systematically solving complex problems involving multiple forces and constraints.
32
If a horse pulls a cart, and the cart pulls back with equal force, **why does the cart move at all**?
The forces **act on different objects**.
## Footnote
Newton's Third Law pairs do not cancel since they act on different bodies. When analyzing the effect of forces using **F** = m**a**, what goes on the left hand side are the forces acting **on** an object.
33
A person in an **elevator feels heavier**. What does this say about the elevator’s motion?
It is accelerating **upward**.
## Footnote
What the person feels is the normal force. Biologically, we do not directly feel the effect of the gravitational force exerted by the Earth since each part of our body experiences the gravitational force in the same way.
Our body can, however, feel the normal force since that only acts on our feet. If we draw the free-body diagram for the person, there is a force mg downwards, and a normal force N upwards.
Assuming that the person moves along with the elevator, the acceleration of the person is the same as the acceleration of the elevator A. Then, by Newton's second law, N - mg = mA, while leads to N = m(g + A). If N is bigger than mg, then A has to be positive (that is, the elevator is accelerating upwards).
34
What is the **acceleration** of a block of mass m pulled with a constant force F at an angle θ on a **frictionless surface**?
## Footnote
Only the horizontal component of the force contributes to acceleration on a frictionless surface. The vertical component of the applied force leads to the normal force no longer being equal to mg.
35
A **heavier object** and a lighter object slide down the same frictionless incline.
Which has greater **speed at the bottom**?
The two objects have the **same speed**.
## Footnote
* Drawing the free-body diagram and applying Newton's second along the incline, mg sin θ = ma.
* m cancels out, meaning that a is the same for both objects (and it's constant).
* Therefore, v will be the same at the bottom as well.
Alternatively, we can use conservation of energy as well to arrive at the same conclusion. Note also that we are assuming that both objects slide - if they roll, then they need not have the same speed at the bottom.
36
# Fill in the blank:
The **normal force** from the floor when a person of mass m stands in an elevator accelerating downward with acceleration a is \_\_\_\_\_.
N = m(g - a)
## Footnote
Drawing the free-body diagram, and applying Newton's second law, we now have N - mg = -mA (minus because the acceleration of the elevator is downwards). So N = m(g - A).
37
# True or False:
The acceleration of a system with two masses m₁ and m₂ connected over a **massless pulley** is a = m₁g / (m₁ - m₂).
False
## Footnote
One way to deduce that this is false is to consider the case m₁ = m₂. With the given expression, the acceleration blows up, which is of course physically nonsense - with two equal masses, we would expect the acceleration to be zero.
To get the correct expression, use Newton's second law for both masses, and assume that the string joining the two masses is also massless (and also inextensible).
For m₁, we write T₁ - m₁g = m₁a₁. For m₂, we have T₂ - m₂g = m₂a₂. Since the pulley is massless, T₁ = T₂, otherwise the pulley would spin infinitely fast.
Also, a = a₂ = -a₁ since the string is inextensible. Now solve simultaneously to get the correct expression.
38
Consider a massless vertical rope supporting a mass m.
What is the **tension in the rope** when the rope is accelerating upwards with acceleration a?
T = m(g + a)
## Footnote
* Draw the free-body diagram for the mass.
* There's mg downwards, and the tension T upwards.
* So T - mg = ma (we assume that the rope does not stretch, so the acceleration of the rope is the same as the acceleration of the mass).
39
A block of mass m initially rests on a **frictionless** incline at angle θ, connected by a massless rope over a massless pulley to a hanging mass 2m. See the figure shown.
Find a formula for the **acceleration of the system** in terms of m, θ, and g.
## Footnote
* Since the pulley is massless, the tension is the same for both masses.
* For the mass on the incline, T - mg sin θ = ma.
* For the vertically hanging mass, 2mg - T = 2ma.
* The acceleration is the same if we assume that the string joining the two is inextensible.
* Add these two equations and simplify.
40
Three blocks A, B, and C are in contact on a frictionless horizontal surface. A horizontal force F is applied to block A (with mass mₐ).
What is the **contact force between blocks B and C**?
## Footnote
* Consider the three blocks together as our system.
* Then the net horizontal force on this system is simply F.
* Assuming that the three blocks move together, they have the same acceleration.
* This common acceleration is then simply F divided by the sum of the masses of the blocks.
* Now look at block C. The only horizontal force on this is the contact exerted by B on C (which is precisely what we want).
* So applying Newton's second law to block C in the horizontal direction, we get the required result.
41
# Define:
Centripetal acceleration
Radially inward acceleration required to **keep an object moving in a circular path**.
## Footnote
It is directed towards the center of the circle.
It is critical in understanding orbits and circular trajectories. As the object moves along a circular arc, its velocity vector keeps on changing direction (even if the magnitude of the velocity vector does not change) - so the object is accelerating. If the object is also changing speed as it is moving in the circular path, then there is a tangential component of the acceleration too.
42
# Fill in the blank:
The **centripetal force** needed to keep an object of mass m moving at speed v in a circle of radius r is given by the formula \_\_\_\_\_.
## Footnote
The magnitude of the centripetal acceleration is v²/r. Applying Newton's second law in the radial direction, we have the desired result. Note that the centripetal force is NOT a separate force - it is simply the name given to the net radially inwards force. For example, for the Earth going around the Sun, there is only the gravitational force acting on the Earth (exerted by the Sun). This gravitational force is the centripetal force in this case.
43
# Fill in the blank:
**Newton's second law** can be expressed in terms of **linear momentum** as the formula \_\_\_\_\_\_.
## Footnote
where p = mv is the **linear momentum**
This form emphasizes the effect of force on changing an object's linear momentum over time, providing a broader perspective on dynamics. This form is especially useful when dealing with systems whose mass is changing (for example, a rocket).
44
A car accelerates **forward**. What force actually propels it?
The **friction force** from the ground on the tires.
## Footnote
Consider the car as the system of particle. For the car to accelerate, there must be an external horizontal force on the car. The only way this can come about is from the friction force exerted by the ground on the tires.
The engine basically causes the tires to rotate; as the tires then push back on the ground, the ground pushes forward on the tires.
45
The work done by the net force acting on an object is defined as W = ∫ 𝐅 ⋅ d𝐬.
What is the physical interpretation of the **dot product** in this context?
The dot product picks up the **component of the force parallel to the displacement**.
## Footnote
Only this component can possibly change the object's speed or kinetic energy.
Without this dot product, we cannot establish the work-kinetic energy theorem.
46
Describe the relationship between **work and energy**.
W = ΔKE
## Footnote
The net work done on an object (defined as the integral of the dot product of the net force on the object with the displacement) is equal to the change in the kinetic energy of the object.
Often, at least some of the work done (the left-hand side of the equation) can be written in terms of changes in potential energy.
47
What are the key **assumptions** necessary for the **conservation of mechanical energy** in a closed system?
* The system is **isolated with no external work** done.
* All forces involved are **conservative**. If they are not, then mechanical energy is not conserved (the total energy is still conserved).
## Footnote
External work is the work done by a force exerted by an object outside the system. If there is external work done, the mechanical energy changes.
These conditions ensure that potential and kinetic energy transform into each other without loss, maintaining constant total mechanical energy. Since the forces involved are conservative, their work done can be expressed in terms of potential energies.
48
When is work considered **negative**?
* Negative work occurs when the force opposes the displacement.
* This follows from the definition of the work.
* Alternatively, if the work done by a force is negative, the force is trying to reduce the kinetic energy.
## Footnote
Negative work is crucial in energy dissipation processes, such as braking or sliding.
Example: Friction acting against the direction of motion for a box sliding on horizontal ground. Since friction points in the direction opposite to the box's displacement, the work done is negative. It results in a reduction of the box's kinetic energy.
49
# True or False:
An object can have **zero work done** on it and still move.
True
## Footnote
If force is zero or perpendicular to the displacement, work = 0. However, the work done is the change in the kinetic energy. If the change in the kinetic energy is zero, that does not, of course, mean that the kinetic energy is zero.
Suppose we have a box sliding on a horizontal frictionless surface with constant velocity. The net work done on the box is obviously zero, and by Newton's first, the box keeps on sliding with the same constant velocity.
50
# True or False:
If a force is **perpendicular** to the displacement of an object, it does no work.
True
## Footnote
Work = ∫ 𝐅 ⋅ d𝐬. If 𝐅 is perpendicular to d𝐬, the dot product is always zero. This makes physical sense: a force perpendicular to the displacement can only change the direction of motion of the object.
It cannot change its speed (or its kinetic energy).
51
# Fill in the blank:
The \_\_\_\_\_ exerted by a force is defined as the **rate of doing work**.
power
## Footnote
Power is mathematically expressed as P = dW/dt. For a constant force, this can be written as P = 𝐅 ⋅ 𝐯.
52
Derive the expression for **gravitational potential energy** in a uniform gravitational field.
## Footnote
This expression is valid near Earth’s surface where g is approximately constant.
The choice of y0 is arbitrary, as only differences in potential energy matter physically. A common choice is 𝑈=0 at ground level or at infinity.
53
How is the **potential energy stored in a spring** related to its displacement?
The potential energy in a spring **increases** as it is stretched or compressed from its equilibrium position.
## Footnote
It depends on how far the spring is displaced, as the more it's stretched or compressed, the more energy it stores.
This potential energy expression is valid for linear elastic springs, where the force is proportional to displacement.
54
Two blocks are pushed across two different surfaces with the **same force** applied over the same distance. One surface is frictionless, the other has friction.
Which block gains more **kinetic energy**?
The block on the **frictionless surface** gains more kinetic energy.
## Footnote
The work done by the applied force is the same in both cases. However, in our scenario, **friction reduces the net work done**. The conclusion is then obvious via the work-kinetic energy theorem.
55
# Fill in the blank:
A **compressed spring launches a block**. If the spring is compressed twice as much, the energy stored increases by a factor of \_\_\_\_\_.
Four
## Footnote
Spring energy is proportional to the square of compression since 𝑈 = 1/2 𝑘𝑥².
56
A person runs up the same flight of stairs twice: **once slowly, once quickly**.
In which case does gravity do more **work**?
Gravity does the **same work** in both situations.
## Footnote
The work done by gravity is independent of the speed of the person.
57
A box of mass m slides down a frictionless ramp of height h and compresses a spring with spring constant k.
How would you find an **expression for the maximum compression x of the spring**?
By using **energy conservation**, the gravitational potential energy of the box gets converted to elastic potential energy of the box-spring system.
## Footnote
We are assuming no loss in the mechanical energy. For example, friction is negligible.
58
A small block of mass goes down the track from an initial height H as shown in the figure. In particular, it goes around a loop.
**What is the smallest value of H such that the block does not fall off the track?**
| Express your answer in terms of the radius R of the loop.
## Footnote
Hint: it must retain contact with the loop!
2.5R
## Footnote
* If the block is able to retain contact at the top of the loop, it is able to retain contact throughout; so let's look at what happens at the top of the loop.
* The normal force is downwards, and so is the weight. We then get N + mg = mv²/R, where R is the radius of the loop.
* Then N = mv²/R - mg. This must be bigger than (or at most equal to) zero - normal forces can push but not pull (the normal force physically cannot flip direction).
* So mv²/R must be bigger than (or equal to) mg. This means that v² must be at least gR at the top of the loop (it cannot be zero!).
* Now apply conservation of energy. Initially the energy is mgH. At the top of the loop, it is (1/2)mgR (at least) plus mg(2R).
* Using conservation of energy, we find that H is at least 2.5R.
59
# Fill in the blank:
The \_\_\_\_\_ \_\_\_\_ \_\_\_\_\_\_ of a system is, on average, the location of the particles comprising a system of particles.
center of mass
## Footnote
This follows from the definition of the center of mass: 𝐑 = (∑ᵢ mᵢ 𝐫ᵢ) / (∑ᵢ mᵢ).
Note that particles with bigger mass are given more 'importance'. Also the center of mass is a position - it is not a mass!
60
# True or False:
In a system of particles, the **center of mass** moves as if all external forces are applied to a point particle (with mass equal to the sum of masses of the particles) located at the center of mass.
True
## Footnote
This follows from the central dynamical equation for a system of particles: 𝐅 = M (d²𝐑/dt²).
The net external force on the system of particles determines how the center of mass moves. This principle allows simplification of complex systems by focusing on the motion of the center of mass, effectively reducing the system to a single particle subject to the net external force.
61
What is the expression for the **velocity of the center of mass** for a two-particle system with masses m₁ and m₂ and velocities 𝐯₁ and 𝐯₂.
## Footnote
Simply start from 𝐑 = (∑ᵢ mᵢ 𝐫ᵢ) / (∑ᵢ mᵢ) for two particles and take the derivative with respect to time on both sides.
Note that the masses of the particles do not change. Note that 𝐕 is a weighted average of the velocities of the two particles (as expected).
62
Explain the **impulse-momentum theorem** for a system of particles.
The **net external impulse equals the change in total momentum** of the system.
## Footnote
* Think of 𝐅 = M (d²𝐑/dt²) as 𝐅 = M (d𝐕 / dt) and integrate both sides with respect to time.
* The left hand side (the integral of the net external force over time) is the impulse.
* The right hand side will give the change in the total linear momentum of the system.
* This theorem is crucial in analyzing collisions and impacts where force may vary over time.
63
Two identical blocks slide on a frictionless surface and are pushed by identical forces. One block is pushed for twice the time.
**Which block has greater kinetic energy?**
The one pushed for **twice** the time.
## Footnote
* Treat the box as the system.
* A greater **impulse (force × time)** implies greater final velocity (due to larger change in linear momentum), and since KE∝v², the kinetic energy is bigger.
* This makes intuitive sense - a longer duration for the applied force means that the box picks up a larger speed.
64
What is the **difference** between **center of mass** and **reduced mass** in a system of two particles?
* The **center of mass** is the position representing the weighted average position of the mass in a system.
* The **reduced mass** is a mass! It simplifies two-body problems by modeling the relative motion as a single particle system.
## Footnote
Center of mass is a **position vector** useful for analyzing external motion; reduced mass is a **scalar** useful for internal interactions like orbits or vibrations.
65
Discuss the role of **energy conservation** in analyzing **collision problems.**
* It helps **determine post-collision velocities**.
* In **elastic collisions**: \( KEinitial = KEfinal \).
* In **inelastic collisions**, energy converts to other forms (such as internal energy), but momentum is always conserved.
## Footnote
Conservation principles simplify complex collision analyses, providing insights into possible outcomes. Remember to define the system of particles in such a way that the net external force is zero - only then is linear momentum conserved!
66
# True or False:
In a **perfectly elastic collision**, kinetic energy is conserved.
True
## Footnote
In perfectly elastic collisions, both kinetic energy and momentum are conserved, unlike inelastic collisions where kinetic energy is not conserved.
67
Two blocks A and B of different masses slide on a frictionless surface and undergo a completely inelastic collision. Before the collision, Block A was moving east while block B was moving north.
After they stick together, in which **direction** do they move?
They move in a direction **between east and north**, determined by the vector sum of the two momenta.
## Footnote
Momentum is a vector quantity; total momentum is conserved in both x and y directions.
Defining the system as the two blocks, the net external force is throughout zero.
Therefore, the total linear momentum is conserved since 𝐅 = M (d𝐕/dt) = d𝐏/dt - if 𝐅 is zero, 𝐏 does not change.
68
Consider a ballistic pendulum. A block of mass M hangs vertically from a string. A bullet of mass m travels horizontally (since bullets usually travel very fast, we ignore the vertical displacement of the bullet) and embeds itself in the block.
How is the **bullet's initial speed u** related to the vertical **height h** by which the bullet and the block go up?
## Footnote
1. Use conservation of linear momentum to find the speed of the block-bullet system just after the bullet embeds itself in the block.
2. Use conservation of energy after the collision.
3. Substitute V and solve for u.
The block and the bullet undergo an inelastic collision, so mechanical energy is not conserved.
69
# Fill in the blank:
In a non-head-on 2D perfectly elastic collision between two identical objects, if one object is **initially at rest**, the two **non-zero final velocity vectors** are separated by \_\_\_\_ degrees.
90
## Footnote
* This is easy to do using vectors.
* Let the mass of each object be m.
* Then, using conservation of linear momentum, m𝐮₁ = m𝐯₁ + m𝐯₂. m cancels on both sides of this equation.
* Take the dot product of this equation with itself. We get u₁² = v₁² + v₂² + 2𝐯₁ ⋅ 𝐯₂.
* Now use conservation of energy since the collision is elastic. Again m cancels here, and we get u₁² = v₁² + v₂².
* 𝐯₁ ⋅ 𝐯₂ = 0.
This is in general only true if the two velocity vectors after the collision are perpendicular to each other.
70
# True or False:
In a one-dimensional **elastic collision** between two objects of equal mass, where one is initially at rest, the moving object comes to rest and **transfers all its speed** to the other object.
True
## Footnote
This happens because both **momentum and kinetic energy are conserved**. It is a typical result in ideal collisions between equal masses, such as in billiard ball interactions.
To see this mathematically (using the usual notation), we have u = v₁ + v₂ using conservation of linear momentum. Since the collision is inelastic, we can use the relationship between the relative velocities.
Doing so, we get u + v₁ = v₂. Solving simultaneously, we immediately get v₁ = 0 and v₂ = 0.
71
Sand falls vertically onto a conveyor belt moving at 2 m/s.
If sand falls at a rate of 0.1 kg/s, **what force is needed** to keep the belt moving at constant speed?
0.2 N
## Footnote
The sand gains horizontal momentum from the belt. To keep moving at constant speed, the belt must exert a force equal to the rate of momentum transfer.
F = (rate) x v = 0.1 x 2 = 0.2N
72
A firework explodes at the top of its arc into two equal-mass pieces. One lands back at the launch point.
**How far does the other piece land from the launch point?**
| (in terms of the original range D)
d = 2D
## Footnote
The center of mass must still land at distance D since internal forces (like the explosion) don't affect its motion.
If one piece lands at 0, the other must land at 2D to keep the average position at D.
73
# Define:
Rigid body
Object that does **not deform under applied forces**; the distance between any two points remains constant.
## Footnote
This idealization allows simplified analysis of motion - for a rigid body, the total work done by the internal forces is zero. This greatly simplifies the analysis via energy methods.
74
# Define:
Angular displacement
**Angle through which a point or line has been rotated** in a specified sense (that is, clockwise or anticlockwise) about a specified axis.
## Footnote
Imagine a thin metallic sheet lying on a horizontal table and we nail this sheet to the table. The sheet can now rotate about the nail. To specify the orientation of the sheet at any instant, we set up a coordinate system with the origin at the nail (this is the simplest choice) and choose a reference point within the sheet.
The angle that the line joining the origin to this reference point makes with the positive horizontal axis (this is the usual conventional choice) gives us the orientation of the sheet in terms of an angle θ (we assume that the sheet is rigid, so this single angle specifies the orientation of the whole sheet). Now, if the sheet rotates, θ changes - we say that we have an angular displacement.
75
What are the **2 key mathematical relationships** in rotational kinematics?
## Footnote
The angular velocity is the derivative of the angular position and the angular acceleration is the derivative of the angular velocity.
Also, the angular velocity is the integral of the angular acceleration, and so on. Consequently, rotational kinematics is quite similar to translational kinematics.
76
What are the **main analogies** between linear and rotational dynamics?
* Position x(t) → Angular position θ(t)
* Linear velocity → angular velocity
* Linear acceleration → Angular acceleration
* Linear momentum → Angular momentum
* Force → Torque
* Mass → Moment of inertia
## Footnote
These analogies facilitate the application of translational mechanics concepts to analyze rotational systems. Note that this is a rather loose dictionary and one needs to be careful.
For example, it is not always true that the angular momentum is equal to the product of the moment of inertia and the angular velocity. Similar, the mass of an object is independent of what the object is doing, but the moment of inertia depends on the axis of rotation.
77
What is the central equation for **rotational dynamics**?
## Footnote
Torque here refers to the net external torque acting on a system of particles, that is, ∑ᵢ 𝐫ᵢ × 𝐅ᵢ, where 𝐅ᵢ is the net external force acting on particle i, and 𝐫ᵢ is any vector that joins the point about which the torque is being calculated to the line of action of 𝐅ᵢ.
Similarly, the total angular momentum is defined as 𝐋 = ∑ᵢ 𝐫ᵢ × 𝐩ᵢ, where 𝐩ᵢ is the linear momentum of the particle i. Note that the angular momentum and the torque have to be calculated with the respect to the same point.
For an object that simply rotating about an axis that is not moving anywhere, this equation simplifies to torque = Iα, where I is the moment of inertia about the axis and α gives the angular acceleration about the same axis.
78
Derive the expression for the **angular momentum** of a system of particles about a fixed point.
## Footnote
This vector sum accounts for each particle's contribution based on its position, mass, and velocity, relative to the fixed point.
79
# Fill in the blank:
\_\_\_\_\_ \_\_\_ \_\_\_\_\_\_ is a measure of an **object's resistance** to changes in rotational motion, depending on mass distribution.
moment of inertia
## Footnote
It plays a similar role to mass in rotational dynamics.
80
# True or False:
The **moment of inertia** of a body is independent of the **axis** about which it rotates.
False
## Footnote
The moment of inertia depends on the **distribution of mass relative to the axis of rotation**. It varies with the choice of axis. This is built into the way that it is calculated: I = ∑ᵢ mᵢ rᵢ². The distances rᵢ depend on the axis of rotation.
81
If a different torque is applied to two objects with equal angular acceleration, which has **greater moment of inertia**?
The one experiencing **greater torque**.
## Footnote
Rearranging τ = Iα, we get that the ratio of the torque to the moment of inertia must be the same for both objects.
82
# Fill in the blank:
The presence of \_\_\_\_\_\_ \_\_\_\_\_\_ leads to the **breakdown of the conservation of angular momentum** in a system.
external torques
## Footnote
This is obvious from the central equation of rotational dynamics.
83
# True or False:
In rotational motion, all points in a rigid body have the same **linear speeds**.
False
## Footnote
In rotational motion about a fixed axis, all points on a rigid body have the same angular velocity ω, but their linear speed v depends on distance r from the axis: v = rw.
84
How does increasing **radius** affect a rotating body's moment of inertia?
It **increases the moment of inertia significantly**, often quadratically.
## Footnote
For most bodies, I ∝ r²; spreading mass farther from the axis increases resistance to rotation. There are exceptions. Consider, for example, the moment of inertia of a cylinder of length L and radius R rotating about an axis passing through its end. In this case, the moment of inertia depends on L and not on R.
85
Determine the expression for the **kinetic energy of a point mass moving in a circular path** about a fixed axis.
## Footnote
For a point mass (m) moving in a circular path of radius (r) with angular velocity (ω), the linear velocity is v = rω.
This is equivalent to the rotational kinetic energy expression \( 1/2 I ω² \) with \( I = mr² \) for a point mass.
86
Which disk has **greater rotational kinetic energy** if two disks of equal radius but different masses rotate with the same angular velocity?
The one with **greater mass**.
## Footnote
Krot = 1/2 Iω², and I increases with mass for the same shape and radius.
87
What condition must be met for **pure rolling** without slipping?
V = Rω
## Footnote
* V is the instantaneous velocity of the center of mass
* R is the radius of the rolling object
* ω is the angular velocity of the rolling object about the center of mass.
The object should translate as much as it rotates. In a time interval Δt, the objects translates a distance VΔt, where V is the velocity of the center of mass.
A point on the edge of the object rotates a distance equal to RωΔt. If these two distances are equal, then V = Rω.
Incidentally, this also means that the velocity of the point of contact with the surface on which the object is rolling is zero - there is a component of the velocity due to the translational motion and another due to the rotational. These two are equal and opposite.
88
What is the **total kinetic energy** for an object rolling without slipping?
The **sum** of both the translational and rotational kinetic energies:
## Footnote
For rolling without slipping, since we have a link between the translational motion and rotational motion, these two terms can generally be combined into a single term.
Rolling motion involves both center-of-mass translation and rotation about the center of mass.
89
# Fill in the blank:
If a bicycle moves without slipping, the **angular velocity of the front wheel (ω₁) and rear wheel (ω₂)** are related through their \_\_\_\_\_.
radii
## Footnote
V is the same for both wheels (otherwise the distance between the wheels would be changing).
Since both wheels are rolling without slipping, we then have ω₁R₁ = ω₂R₂, so ω₁/ω₂ = R₂/R₁.
Angular velocities are inversely proportional to wheel radii.
90
Two solid spheres roll from rest down an inclined plane. One has twice the mass and radius of the other.
**Which reaches the bottom first, and why**?
The spheres reach the bottom at the **same time** because the acceleration depends only on the geometry, not mass or size.
## Footnote
For solid spheres, a= (5/7) gsinθ, which doesn’t depend on mass or radius. This comes from combining Newton’s laws and the condition for rolling without slipping. Static friction enables rotation, but does no net work.
91
What happens to a figure **skater's angular velocity** when she pulls in her arms during a spin?
It **increases,** due to conservation of angular momentum.
## Footnote
L = Iω in this case; decreasing I leads to increasing ω.
92
# Define:
Oscillatory motion
Repetitive motion about an **equilibrium point**, such as a pendulum or spring.
## Footnote
This type of motion is often periodic and can be described using sine or cosine functions.
93
# True or False:
In **simple harmonic motion (SHM)**, the acceleration of the particle is directly proportional to its displacement from the **equilibrium position** and directed towards the equilibrium position.
True
## Footnote
In SHM the **restoring force** is proportional to displacement and directed toward equilibrium.
Mathematically, this leads to the differential equation that the double derivative of a position (or angle) is proportional to the position (or angle), with a negative sign. The solutions of this differential equation are oscillatory.
94
In a simple harmonic oscillator, how are the phases of **displacement**, **velocity**, and **acceleration** related to each other?
* Displacement and velocity are **90 degrees out of phase.** This is because the derivative of the sine function is the cosine function, and the derivative of the cosine function is the sine function.
* Displacement and acceleration are **180 degrees out of phase**. This follows from the fact that the acceleration is proportional to the displacement (with a negative sign).
* Velocity and acceleration are **90 degrees out of phase.** Consider the previous two points.
## Footnote
These phase differences are crucial for understanding energy transfer in oscillatory systems.
95
If the **frequency of a mass-spring oscillator is doubled**, what happens to the period?
It becomes **half**.
## Footnote
Frequency and period are reciprocals: f = 1/T.
96
# True or False:
Doubling the **amplitude of a spring-mass oscillator** doubles its period.
False
## Footnote
Period in simple harmonic motion is independent of amplitude. This follows from the differential equation. Here we also assume that Hooke's law continues to remain valid even with the larger amplitude.
97
Two mass-spring systems oscillate with the same amplitude but different spring constants.
Which system has the **shorter period**?
The one with the **larger spring constant**.
## Footnote
The restoring force is -kx.
Applying Newton's second law and rearranging, we find that the double derivative of x is equal to -(k/m)x. We then identify the square of the angular frequency to be (k/m).
This leads to T=2π√(m/k); larger k → smaller period.
98
# True or False:
The **period of a pendulum** depends on its mass.
False
## Footnote
T=2π√(L/g); mass does not appear in the formula.
One quick way to derive this is to look at the restoring torque -mgl sinθ and set this equal to Iα with I = ml².
* Make the small angle approximation to replace sinθ ≈ θ.
* Rearrange to get the angular frequency and hence the time period.
Physically, a larger mass increases the restoring torque for the simple pendulum (since the weight is bigger), but the inertia also increases. These two effects cancel each other out.
99
Pendulum B has a period **three times** that of pendulum A. How do their **lengths compare**?
The length of B is **9 times** the length of A.
## Footnote
The period of a pendulum is proportional to the square root of its length.
100
# True or False:
A spring-mass system that **oscillates on the Moon (lower g)** has a different period than on Earth.
False
## Footnote
T=2π√(m/k); gravity doesn’t affect mass-spring oscillators.
That's simply because the restoring force in this case is essentially due to the spring force (even if the spring is vertical) rather than gravity. All gravity does for vertical springs is to effectively change the equilibrium position.
101
Which change will **decrease the frequency** of a spring-mass oscillator?
Increase mass
## Footnote
f=1/(2π√(m/k)); more mass → lower frequency.
102
# Fill in the blank:
A **simple harmonic oscillator** has a maximum speed when it passes through its \_\_\_\_\_\_ \_\_\_\_\_\_.
Equilibrium position
## Footnote
At equilibrium, all energy is kinetic. Potential energy is zero.
103
# Fill in the blank:
A **simple pendulum** has a period T. If its **length is quadrupled**, the new period is \_\_\_\_.
2T
## Footnote
The period of a pendulum depends on length as 𝑇∝√𝐿
104
An object is attached to a **horizontal spring** and oscillates with angular frequency ω. If the **spring constant** is doubled and the mass remains the same, what happens to ω?
ω **increases** by a factor of √2
## Footnote
Angular frequency of a spring-mass system is ω= √k/m
105
# Fill in the blank:
The energy in a **simple harmonic oscillator** is proportional to the \_\_\_\_\_\_ squared.
amplitude
## Footnote
The restoring force is proportional to the displacement from equilibrium. So the potential energy is proportional to the square of the displacement.
When the object is maximally displaced, the kinetic energy is zero and the total energy is simply the potential energy.
Therefore, the total energy E∝A²; higher amplitude → more energy.
106
At what points in the motion of a **block attached to a spring** are the kinetic and potential energies equal?
When the **displacement** is x = ±A/√2.
| (𝐴 is the amplitude.)
## Footnote
* Use the expressions for **kinetic and potential energy in SHM**: 𝐾 = 1/2 𝑘(𝐴² − 𝑥²), 𝑈 = 1/2 𝑘𝑥².
* Set 𝐾 = 𝑈 and solve for 𝑥.
Note that in simple harmonic motion problems, we usually find the kinetic energy by first finding the total energy and then the potential energy.
107
Describe how **energy transformations** occur in a **simple pendulum**.
* At maximum displacement, energy is entirely potential since the kinetic energy is zero.
* At the lowest point, energy is entirely kinetic.
* **Energy oscillates between kinetic and potential forms** as the pendulum swings.
## Footnote
The simple pendulum exemplifies periodic energy transformation in conservative systems.
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