Momentum

The linear momentum of a particle is a vector quantity obtained by multiplying the mass of an object by its velocity.

\[ \vec{p} = m\vec{v} \]

Momentum has units kg m/s or Newton seconds ( N s)

Newton's second law states that the rate of change of momentum of a body is directly proportional to the force applied, and this change in momentum takes place in the direction of the applied force.

Newton's 2nd Law

\[ \mathbf{F} = \frac{d\mathbf{P}}{dt} \] \[ = \frac{d(m\mathbf{v})}{dt} \] \[ = \frac{m\,d(\mathbf{v})}{dt} \] \[ = m\mathbf{a} \] \[ \text{since} \frac{d(\mathbf{v})}{dt} = \mathbf{a} \]

A force acting on a system affects its momentum.

If there is no net external force, there is no change in momentum.

If an object changes velocity, its momentum will change.

Example

Elephant momentum example

An elephant with a mass of 5,000 kg changes his velocity from 2 m/s to 4 m/s in the same direction. What is the change in his momentum ?

\[ \vec{P} = m\vec{v} - m\vec{u} \] \[ = 5000 \times 4 - 5000 \times 2 \] \[ = 20000 - 10000 \] \[ = 10000\ \text{Ns} \]

\[ \Delta p = m(v - u) = 5000(4 - 2) = 10000\text{ kg·m/s} \]

Impulse

Impulse ( J )occurs when a constant force acts over a time interval.

so Impulse = Force x Time

\[ \mathbf{J} = \int_{\Delta t} \mathbf{F}\,dt \] \[ = \int_{t_1}^{t_2} \mathbf{F}\,dt \] \[ = \int_{t_1}^{t_2} m\mathbf{a}\,dt \] \[ = m \int_{t_1}^{t_2} \mathbf{a}\,dt \] \[ = m \bigl[\mathbf{v}\bigr]_{t_1}^{t_2} \] \[ = m\mathbf{v}(t_2) - m\mathbf{v}(t_1) \]

Impulse is the change in momentum.

\[ J = Ft \] \[ J = \Delta p \]

Example

A force of \( (5\mathbf{i} + 10\mathbf{j}) \) newtons acts on a body of mass 10 kg for 4 seconds.
The body was initially moving with a constant velocity of \( (2\mathbf{i} - 7\mathbf{j}) \) m s^-1.
Find the final velocity of the body in vector form, and hence obtain its final speed.

\[ \text{Impulse} = \text{Force} \times \text{time} \] \[ 10\mathbf{v} - 10(2\mathbf{i} - 7\mathbf{j}) = (5\mathbf{i} + 10\mathbf{j}) \times 4 \] \[ 10\mathbf{v} = 20\mathbf{i} + 40\mathbf{j} + 20\mathbf{i} - 70\mathbf{j} \] \[ 10\mathbf{v} = 40\mathbf{i} - 30\mathbf{j} \] \[ \mathbf{v} = \frac{40\mathbf{i} - 30\mathbf{j}}{10} \] \[ \mathbf{v} = 4\mathbf{i} - 3\mathbf{j} \]

\[ \text{speed} = |v| \] \[ = |\,4\mathbf{i} - 3\mathbf{j}\,| \] \[ = \left|\sqrt{4^2 + (-3)^2}\right| \] \[ = \left|\sqrt{25}\right| \] \[ = 5\ \text{m s}^{-1} \]

Conservation of Momentum

Newton's 3rd Law

Newton’s Third Law implies:

\[ \vec{F}_A + \vec{F}_B = 0 \]

Therefore total momentum is conserved:

\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]

In an elastic collision, both momentum and kinetic energy are conserved.

In an inelastic collision, only momentum is conserved. Some kinetic energy is transferred into another form, e.g. heat or sound.

In a completely inelastic collision, the objects stick together.

Some scenarios

1) A ball hits a stationary ball of equal mass.

collision occurs

one ball remains stationary

2) A ball hits a lighter stationary ball.

collision occurs

both move

3)A ball hits a lighter stationary ball.

collision occurs

the balls stick together

Example

A bullet of mass 20 grams, travelling at 80 ms^-1, hits a stationary block of wood of mass 4 kg which is free to move on a smooth, horizontal plane. Use the conservation of linear momentum to calculate the final speed of the block, given that the bullet passes right through the block and emerges with speed 20 m s^-1.

\[ m_1\mathbf{u}_1 + m_2\mathbf{u}_2 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2 \] \[ 0.02 \times 80 + 4 \times 0 = 0.02 \times 20 + 4\mathbf{v}_2 \] \[ 4\mathbf{v}_2 = 1.6 - 0.4 \] \[ \mathbf{v}_2 = \frac{1.2}{4} \] \[ \mathbf{v}_2 = 0.3\ \text{m s}^{-1} \]

Speed is the modulus of velocity, so the speed is 0.3 ms^-1

Example

A body (E) of mass 6 kg is moving with velocity (−2i+5j) ms^-1when it collides with a body (F) of mass 10 kg moving with velocity (3i+6j) ms^-1. Immediately after the collision the velocity of E is (3i+2j) ms^-1. Find the velocity , v, of F immediately after the collision.

before

after

\[ (m\mathbf{v})_{\text{before}} = (m\mathbf{v})_{\text{after}} \] \[ 6(-2\mathbf{i} + 5\mathbf{j}) + 10(3\mathbf{i} + 6\mathbf{j}) = 6(3\mathbf{i} + 2\mathbf{j}) + 10\mathbf{v} \] \[ -12\mathbf{i} + 30\mathbf{j} + 30\mathbf{i} + 60\mathbf{j} = 18\mathbf{i} + 12\mathbf{j} + 10\mathbf{v} \] \[ 18\mathbf{i} + 90\mathbf{j} = 18\mathbf{i} + 12\mathbf{j} + 10\mathbf{v} \] \[ 10\mathbf{v} = 78\mathbf{j} \] \[ \mathbf{v} = 7.8\mathbf{j} \]

Hooke’s Law

A force acts to restore a spring to its relaxed state, where it is neither extended nor compressed.

Robert Hooke discovered that the force from a spring is proportional to the displacement of the free end from its relaxed state position.

\[ F \propto x \] \[ F = kx \]

The constant, k, is called the spring constant and measures the stiffness of the spring.

Relaxed state - no force.

Compressed state

The spring has been compressed x units to the left.

Since x is negative, the force experienced is positive.

Extended state

The spring has been compressed x units to the right.

Since x is positive, the force experienced is negative.

When used as a restoring force, Hooke's Law can be written

\[ F = -kx \]

Hooke's Law can also be expressed as

\[ T = \frac{\lambda x}{l} \]

Where T is the tension, x is the extension, l is the natural length of the spring and λ is the modulus of elasticity of the string.

Work Done

Work is done when there is displacement of a point of application of a force in the direction of the force.

Work done by a constant force

\[ W = \vec{F}\cdot\vec{s} \]

Work is scalar, with SI units Joules (J), yet Force and displacement are vectors.

\[ W = Fs \] \[ = mas \] \[ = m\left(\frac{v^2 - u^2}{2}\right) \] \[ = \frac12\left(mv^2 - mu^2\right) \]

The work done on an object by a net force equals the change in kinetic energy of the object.

\[ W = \Delta E_K \]

Power

Power is the rate of doing work:

If a force F does work W during a time interval Δt, then the average power due to the force over the time scale is

\[ P_{\text{avg}} = \frac{W}{\Delta t} \]

At any particular point of time ,

\[ P = \frac{dW}{dt} \] \[ = \frac{d(Fs)}{dt} \]

but the force is constant and \(\frac{ds}{dt} = v \) , so

\[ P = Fv \]

If the direction of the force is at an angle \( \theta \) to the direction of travel of the object,

\[ P = Fv\cos\theta \]

then instantaneous power is

\[ P = \vec{F}\,\cdot\,\vec{v} \]

Example

Doggo decided to be lazy and accepted a lift from a pleasure boat.

The tow rope exerts a force of 50 N on the kayak at an angle 60˚ to the horizontal.

If the instantaneous power is 100 W, what is the magnitude of the velocity of the kayak ?

\[ P = \vec{F}\cdot\vec{v} \] \[ p = |F|\,|v|\cos\theta \] \[ 100 = 50\cos 60^\circ \, |v| \] \[ |v| = \frac{100}{50\cos 60^\circ} \] \[ |v| = 4 \] \[ \text{The velocity of the kayak is } 4\text{ m/s} \]

Kinetic Energy

For a body of mass m moving at a speed (or magnitude of velocity) v,

\[ \text{Kinetic Enery }= \frac{1}{2}mv^2 \]

\[ E_K= \frac{1}{2}mv^2 \]

Using momentum:

\[ P = mv \] \[ P^2 = (mv)^2 = m^2 v^2 \] \[ E_k = \tfrac12 mv^2 \] \[ = \frac{mv^2}{2} \] \[ = \frac{m^2 v^2}{2m} \] \[ = \frac{P^2}{2m} \]

\[ E_K= \frac{p^2}{2m} \]

Kinetic Energy is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

\[ W = Fs = \frac{1}{2}mv^2 \]

Giving

\[ \frac{2Fs}{m} = v^2 \] \[ v = \sqrt{\frac{2Fs}{m}} \] \[ v = \sqrt{\frac{2W}{m}} \] \[ \frac{2Fs}{v^2} = m \] \[ \frac{2W}{v^2} = m \]

Potential Energy

\[ E_P = mgh \] \[ \text{Or} \] \[ U= mgh \]

The change in potential energy is \( \Delta U = -W \)

where the negative value shows tht work done against a force increases potential energy.

For one dimensional motion,

\[ F(x) = -\,\frac{dU(x)}{dx} \]

Conservation of Energy

In a system where no work is done against friction and gravity is the only external force which does work on a body :

\[ KE_1 + PE_1 = KE_2 + PE_2 \]