Newton's Laws of Motion and Linear Momentum

Newton’s three(3) laws of motion are as follows:

1st law (the law of inertia)

A body in motion, or at rest, remains in its state of motion or rest, unless acted upon by an external unbalanced force. Bodies with more mass are more resistant to changes in motion thus mass is a direct measure of a body’s inertia (its ability to resist changes in motion).

The same force that causes a small change in motion for a heavy object (e.g. a boulder) will cause a large change in motion for a lighter object(e.g. a ping pong ball).

2nd law

The rate of change of momentum of a body is directly proportional to the force applied to it and this change takes place in the direction of the applied force. The rate of change of momentum can be written as $\frac{mv-mu}{t}$ (the difference in final momentum and initial momentum over time). Thus the proportionality can be written as: $$F\propto\frac{mv-mu}{t}$$ Because the acceleration, $a=\frac{v-u}{t}$: $$F\propto ma$$ The constant of proportionality is $1$: $$F=ma$$

3rd law

Every action has an equal and opposite reaction. When a force is exerted onto an object, it will exert a reaction force to counter that original force.

Linear momentum

The linear momentum($p$) of a body is the product of its mass and its velocity: $$p=mv$$

This formula implies that the SI unit for linear momentum is: $$ \begin{equation}\begin{aligned} p=mv\rightarrow kg\times ms^{-1}=kgms^{-1}\\ \end{aligned}\end{equation} $$

The bigger the body, the more momentum it has. The faster a body is moving, the more momentum it has. A small object travelling at a high velocity can have a much greater momentum than a larger object moving at a lower velocity.

The Principle of Conservation of Linear Momentum

The total momentum of a system before a collision is equal to the total momentum after the collision, given that no external forces are acting.

$$ \begin{equation}\begin{aligned} m_1v_1+m_2v_2 = m_1v_3 + m_2v_4\\ \end{aligned}\end{equation} $$

Example 1

A stationary ball B mass $5\ kg$ is struck by another ball A of mass $2\ kg$ moving at $4\ ms^{-1}$. Find the final velocity of B if Ball A continues in the same direction as ball B but with velocity $1\ ms^{-1}$.

What we have:

• $m_1=2kg$
• $m_2=5kg$
• $v_1=4ms^{-1}$
• $v_2=0ms^{-1}$
• $v_3=1ms^{-1}$
• $v_4=?$

$$ \begin{equation}\begin{aligned} total\ momentum\ before&=total\ momentum\ after\\ m_1v_1+m_2v_2&=m_1v_3+m_2v_4\\ m_2v_4&=m_1v_1+m_2v_2-m_1v_3\\ v_4&=\frac{m_1v_1+m_2v_2-m_1v_3}{m_2}\\ v_4&=\frac{2kg\times 4ms^{-1}+5kg\times 0ms^{-1}-2kg \times 1ms^{-1}}{5kg}\\ v_4&=1.2ms^{-1}\\ \end{aligned}\end{equation} $$

Example 2

Find the final velocity of B if Ball A is brought to rest by the collision.

What we have:

• $m_1=2kg$
• $m_2=5kg$
• $v_1=4ms^{-1}$
• $v_2=0ms^{-1}$
• $v_3=0ms^{-1}$
• $v_4=?$

$$ \begin{equation}\begin{aligned} total\ momentum\ before&=total\ momentum\ after\\ m_1v_1+m_2v_2&=m_1v_3+m_2v_4\\ m_2v_4&=m_1v_1+m_2v_2-m_1v_3\\ v_4&=\frac{m_1v_1+m_2v_2-m_1v_3}{m_2}\\ v_4&=\frac{2kg\times 4ms^{-1}+5kg\times 0ms^{-1}-2kg \times 0ms^{-1}}{5kg}\\ v_4&=1.6ms^{-1}\\ \end{aligned}\end{equation} $$

Example 3

Find the final velocity of B if Ball A rebounds in the opposite direction with a velocity of $2\ ms^{-1}$.

What we have:

• $m_1=2kg$
• $m_2=5kg$
• $v_1=4ms^{-1}$
• $v_2=0ms^{-1}$
• $v_3=-2ms^{-1}$
• $v_4=?$

$$ \begin{equation}\begin{aligned} total\ momentum\ before&=total\ momentum\ after\\ m_1v_1+m_2v_2&=m_1v_3+m_2v_4\\ m_2v_4&=m_1v_1+m_2v_2-m_1v_3\\ v_4&=\frac{m_1v_1+m_2v_2-m_1v_3}{m_2}\\ v_4&=\frac{2kg\times 4ms^{-1}+5kg\times 0ms^{-1}-2kg \times -2ms^{-1}}{5kg}\\ v_4&=2.4ms^{-1}\\ \end{aligned}\end{equation} $$

Example 4

Ball C with mass $4\ kg$ moving left at $5\ ms^{-1}$ collides with ball D mass $2\ kg$ moving right at $8\ ms^{-1}$. What is the final velocity of C if D rebounds left with velocity $4\ ms^{-1}$?

Example 5

A bullet of mass $20\ g$ is fired from a gun with a muzzle velocity of $100\ ms^{-1}$. The mass of the gun is $5\ kg$. Find the recoil velocity of the gun.

Elasticity of a collision

An elastic collision is one where the total kinetic energy before and after the collision are the same. Therefore, an inelastic collision sees the total kinetic energy differing before and after the collision.

Linear momentum is always conserved, regardless of if we observe an elastic or inelastic collision. The only factor that can determine if linear momentum is not conserved is the presence of an external force.

Consider example 3 above:

What we know:

• $m_1=2kg$
• $m_2=5kg$
• $v_1=4ms^{-1}$
• $v_2=0ms^{-1}$
• $v_3=-2ms^{-1}$
• $v_4=2.4ms^{-1}$

We can find the initial and final kinetic energies of the balls before and after the collision: $$ \begin{equation}\begin{aligned} KE_{i(A)} &= \frac12 m_1v_1^2\\ &=\frac12 (2kg)(4ms^{-1})^2\\ &=16J\\ KE_{i(B)} &= \frac12 m_2v_2^2\\ &=\frac12 (5kg)(0ms^{-1})^2\\ &=0J\\ KE_{f(A)} &= \frac12 m_1v_3^2\\ &=\frac12 (2kg)(-2ms^{-1})^2\\ &=4J\\ KE_{f(B)} &= \frac12 m_1v_4^2\\ &=\frac12 (5kg)(2.4ms^{-1})^2\\ &=14.4J\\ \end{aligned}\end{equation} $$

Since the total kinetic energy before ($16J+0J=16J$) does not match the total kinetic energy after ($4J+14.4J=18.4J$), we conclude that the collision is inelastic.