Revision Note 7: Momentum and Collisions

Momentum:

p = mv

What is important is change in momentum,

F∆t = ∆p = m∆v

Impulse:

F∆t is called the impulse, which is equal to change in momentum.

Use calculus when force varies (generally with time):

Center of mass:

For n different masses, the center of mass (x,y) is:

Problem Solving Tips:

Tip 7.1:

For solving momentum problems, first resolve momentum into two orthogonal axes. Then:

p_i,x = p_f,x

p_i,y = p_f,y

The above apply to all types of momentum problems.

The issue is with the number of unknowns and number of equations. There are now three types:

(a) Apply the principle of conservation of momentum only. Some parts of the final momentum are given – maybe the angle or the magnitude – to provide the extra equations to enable solution.

(b) Completely Elastic. No energy lost.  In this case we have the extra equation:

KE_i = KE_f

Or if springs are involved:

KE_i + PE_i = KE_f + PE_f

(c ) Completely inelastic. No, not all energy is lost, but some definitely is lost. The definition of a completely inelastic collision is that the two objects stick together after collision, so their final velocities are the same:

v_1,f = v_2,f

Broken down into two components:

v_x,1,f = v_x,2,f

v_y,1,f = v_y,2,f

Revision Note 6: Work and Energy

Work done

Work done by a single force, F, which applies through a displacement, d, is given by the scalar product:

W = F.d = Fd cosθ

You need calculus if any one of the following is true:

·        The magnitude of the force varies (generally as a function of distance)

·        The θ varies (generally as a function of distance)

In the Calculus version, add up the work done by the force for small distances (dx) – which requires integration from x = x_i = initial position to x = x_f = final position. In other words:

If multiple forces act on the object then the total work done by all forces together is:

·        Algebraic sum of the work done by each force individually.

·        Work done by the net force.

·        Sum of work done over small distances covering the whole distance (integration method)

Work and Energy

If A applies a force on B and does W work, then B does  –W work on A. This comes from Newton’s third law. The object A also transfers W amount of energy to B, and the object B transfers  –W of energy to A.

Work energy theorem:

W_{all forces} = ∆K

Potential energy theorem:

W_grav = -∆U

If the only force is gravitation (or any conservative force) then W_grav = -∆U = ∆K. In this case,

∆K + ∆U = 0

or K + U = constant, which means for initial and final positions:

K_i + U_i = K_f + U_f

The intuition of conservative forces is that work doesn’t get lost and only gets transferred between two bodies – hence talking about potential energy makes sense. PE won’t make sense in case energy leaks out.

Potential Energy

A natural way to measure gravitation potential energy is to set the gravitational PE to 0 when the separation of bodies is infinity. In that case:

Similar considerations apply to energy of an electron around the nucleus and thus that electron while in orbit has negative energy.

Potential energy of a spring of spring constant k compressed by x is given by:

Finding F_grav(x) from U(x):

Power

If P is the power (scalar), then for a constant force, F:

Problem Solving Tips:

Tip 6.1:

For computing W, since cosθ = cos(-θ) we don’t care if the angle is measured from Force to displacement or vice-versa – we just say that it is the angle between the Force and displacement vectors.

However the work done can be negative or zero – see the following diagram:

Scan It

Tip 6.2:

For work, what is important is the displacement, not the distance (though see Work done by friction below). So work done by a constant force through a round trip is zero since displacement is 0.

Tip 6.3:

For work done by friction, the magnitude of the force F_fr = μN continually adjusts to oppose the motion, so θ=180 always.  So for each dx the work done is W_fr = -F_fr.dx. Total work is W_fr = -F_fr.D where D = distance covered. In this case work depends on distance rather than displacement.

Tip 6.4:

When conservative forces are acting, the work done doesn’t depend on path. The potential energy is dependent on location, not the path taken to it.

Tip 6.5:

The spring has the same potential energy for the same amount of compression or same amount of extension. This is because (x)² = (-x)².

Tip 6.6:

Work done by conservative forces = difference in the potential energy between starting and ending locations.

Revision Note 5: Friction

Frictional force (more correctly, the maximum frictional force) is normal force times coefficient of static or dynamic friction (as the case may be).

F_fr = μN

The normal force (N) that is used to compute frictional force is best computed by considering the force by the surface on the body – in other words you need the free-body diagram of the body. This is the best method to compute the force by the body on the surface, as by Newton’s third law the two forces are numerically the same.

Problem Solving Tips:

Tip 5.1:

What is F_fr  for a ramp of incline θ to horizontal when the coefficient of friction is μ? By a previous tip (Tip 2.3), N = mg.cosθ. So,

F_fr,θ = μN = μ mg.cosθ

Tip 5.2: Up-down intuition:

Relating θ with normal force, N, and frictional force, F_fr,θ:

θ ↑        N ↓      F_fr,θ  ↓

The effect of friction decreases as the ramp becomes more and more vertical.

Revision Note 4: Newton’s Second Law

Newton’s Second Law:

F_net = ma

Here,

·        F_net = the net force (sum of all forces) on the body

·        m = the mass of the body

·        a = the acceleration of the body

When F_net = 0, a = 0 Newton’s second law becomes equivalent to his first law.

Problem Solving Tips:

Tip 4.1:

The following diagram relates the techniques in Revision Notes 2 (Static equilibrium), Revision Note 3 (Kinematics) and Revision Note 4 (Newton’s second law):

Tip 4.2:

When multiple bodies are pulled in a chain (see picture), then Newton’s second law is satisfied by (i) both bodies together, (ii) Body A, (iii) Body B.

In the example below, F = 2N, T = 1N. Where did the extra 1N force go? It goes to feed the acceleration of Body B. The ‘rough’ statement of this:  Acceleration eats Force.

Tip 4.3:

Impulse is defined as F.∆t, and that is equal to change in momentum:

F.∆t = ∆p

Tip 4.4:

These vectors have the same directions: F, ∆p, ∆v, and that direction is unrelated to the directions of v, p. Note that ∆v and v may not have the same direction. In Physics, v is not so important, ∆v is.

Revision Note 3: Kinematics

The following three equations of kinematics are valid only when acceleration is constant:

1.      v(t) = v(0) + at

2.      x(t) = x(0) + v(0)t + ½ at²

3.      [v(t)]² = [v(0)]² + 2a[x(t) – x(0)]

Here:

·        Time = t

·        Acceleration = a

·        Velocity at time t = v(t)

·        Position at time t = x(t)

Motion in 2D
The x- and y-directions have separate kinematics equations that do not interact. Only time connects the two.  So, the position of the particle at time t is (x(t), y(t)). If the particle is stopped (e.g., by falling to the ground), then the time of flight is the time when it hits the ground – generally this comes from the vertical kinematics equations.

If one direction (generally horizontal = x-direction) doesn’t have any acceleration, then the distance travelled in that direction is easy to find (x = vx . t). The total distance travelled in that direction then is velocity times time of flight.

Problem Solving Tips:

Tip 3.1:

When the acceleration is not constant it is expressed as a(t). Then use Calculus to find v(t), x(t) etc by integrating a(t).

Tip 3.2:

The time for a falling body to drop a height, h, is given by h = ½ gt².

Tip 3.3:

If you throw a particle vertically upwards with velocity, v, then the maximum height it reaches is given by the equation: v² = 2gh. This equation can be derived via kinematics and also from energy principles:

PE_i    +      KE_i       =    PE_f     +   KE_f

0       +   ½ mv²   =    mgh   +    0

        v²   =  2gh

Tip 3.4: What goes up …

What goes up in time, t, also comes down in time, t. In other words the equation from Tip 3.2 works in both directions – going up and going down. Not just that, the velocities going up and going down at any point in the flight have the same magnitude (the directions are opposites of each other).

Tip 3.5: Inclined planes

A particle pushed up an inclined plane (without friction) gains the same vertical height as a particle thrown vertically up. This follows from the energy principles as in Tip 3.3.

Even though the vertical heights are the same, the particle on the inclined plane travels a s longer distance and takes more time.

Tip 3.6: Up-down intuition:

Relating the height, h, from which a ball is dropped with the time, t, of flight:

h ↑ t ↑

The time goes up is less since for the latter part of the journey the speed of the ball is very high.

Relating the velocity, v, of a ball thrown up and the height, h:

v ↑ h ↑

Here too higher initial velocity cannot boost the height too much since the latter part of the journey is covered at lower speeds.

Tip 3.7: Projectile motion:

If v = initial velocity and θ = angle of launch (projectile launched from the ground level), then:

Maximum height = h = v².sin²θ / 2g

Range = R = v².sin 2θ / g

Time of flight = T = 2v.sin θ /g

The maximum value for h, T are achieved when θ=90 and for R when θ=45.

For maximum range: v_x = v_y at launch.

Revision Note 2: Static equilibrium

Conditions for static equilibrium:

1.       F_net = 0 (net force is zero)

2.       τ_net = 0 (net torque is zero)

For calculations:

·         Break F_net down into F_x, F_y (with proper signs) and then test for zeros along each axis.

·         For rotational moments, balance the clockwise and anti-clockwise moments.

Problem Solving Tips:

Tip 2.1:

In a force diagram, you will have to decide which direction is positive and which is negative. You can choose anything, but it helps to decide beforehand so that you do not waste time during the exam debating this.

Convention:

Vertically up is positive: E.g., g = -9.8 ms^-2.

Horizontally right is positive.

Along the Slope: x-axis, positive to the right.

Perpendicular to the slope: y-axis, positive upwards.

Tip 2.2:

Unknowns and signs: Suppose there is an unknown tension, T.

First, if you know its expected direction, draw it that way. Then use T or -T depending on the direction you chose. If you guessed the direction right, the numerical answer will be positive. If the numerical answer is negative, then the actual direction is in the reverse. It is like this: acceleration a = -5ms^-2 rightwards is actually +5ms^-2 leftwards.

Tip 2.3:

If θ is the angle of the inclined plane and a particle of mass m is on it, then the component of the weight along the x-axis and the y-axes are mg.sin θ and mg.cos θ respectively. Use the principle of extreme values to check this answer – when θ = 0 (plane is flat), then the component along the plane is 0 as it should be. This may seem counter to the “shadow = cos θ” tip, but is not since the θ is different. The correct angle (90 - θ) would match up with the shadow tip.

Tip 2.4: Up-down intuition:

Relating θ of inclined plane to a_x = component of g along the plane

θ ↑ a_x ↑

Intuition: When θ=90 then a_x = g, the highest possible value.

Revision Note 1: Vectors and components of vectors

There are two ways to represent a 2D vector, v:

1.       (v, θ)

·         v is the magnitude of the vector v.

·         θ is the angle that the vector’s direction makes with the x-axis.

2.       v = v_xi + v_yj

·         v_x is the x-component of the vector.

·         v_y is the y-component of the vector.

The second representation is more convenient for problems. An intuitive connection between the two representations is this: The vector v can be represented by an arrow drawn from (0, 0) to (v_x, v_y). In that case the length of the arrow will be v and it will make an angle θ with the x-axis.  

Problem Solving Tips:

Tip 1.1:

Drawing vectors: Put the tail of the arrow at the point of application – e.g., the arrow representing a force vector, F, should have the tail on the body on which F is applied. A common mistake (especially for force vectors) is to put the head at the point of application.

Tip 1.2:

Going from (v_x, v_y) form to (v, θ) form:

Tip 1.3:

The component of a vector v along a certain direction is v.cos θ where θ is the included angle between the direction of the vector and the direction of the component.

This is the “shadow = cos θ” tip. (shadow = the component of the vector with sun at noon.)

The component perpendicular to the above is v.cos (90-θ) = v.sin θ.

Tip 1.4:

The component of a vector along a direction perpendicular to itself is 0. This principle of orthogonality is used to simplify problems.

Tip 1.5: Up-down intuition:

As the angle θ that the vector makes with the horizontal increases from  to , the horizontal component v_x decreases. The arrows in the notation represent this relationship. The relationship v_x = v.cos θ is not exactly linear.

θ ↑ v_x ↓