Force and Laws of Motion • Topic 2 of 3

Newton's Second Law & Momentum

Momentum. The momentum of a moving body is the product of its mass and its velocity. It tells us how much motion a body has and how hard it is to stop. Momentum is written as $p = mv$, where $m$ is the mass in kilograms and $v$ is the velocity in metres per second. Momentum is a vector quantity, pointing in the same direction as the velocity. Its SI unit is $\text{kg m/s}$ (kilogram metre per second). A heavy truck moving slowly and a light bullet moving fast can both carry large momentum, which is why both are hard to stop.

Newton's Second Law. The rate of change of momentum of a body is directly proportional to the applied unbalanced force and takes place in the direction of the force. In symbols, $F \propto \frac{\Delta p}{\Delta t}$. If a force $F$ acts for time $t$ on a body whose velocity changes from $u$ to $v$, then $\Delta p = mv - mu$ and $F = k\frac{\Delta p}{\Delta t} = \frac{m(v - u)}{t}$.

Deriving $F = ma$. Since acceleration $a = \frac{v - u}{t}$, we get $F = ma$ (taking the constant $k = 1$). This is the most useful form of the Second Law. It shows that:

  • For a fixed mass, a larger force produces larger acceleration.
  • For a fixed force, a larger mass produces smaller acceleration.

The newton defined. The constant $k = 1$ fixes the SI unit of force. One newton is the force that gives a mass of $1\ \text{kg}$ an acceleration of $1\ \text{m/s}^2$. So $1\ \text{N} = 1\ \text{kg m/s}^2$. Newton's First Law is actually a special case of the Second: when $F = 0$, $a = 0$, so the velocity stays constant.

Why force depends on time of contact. Because $F = \frac{\Delta p}{\Delta t}$, the same change in momentum needs a smaller force if the time is longer. This explains many real situations:

  • A cricketer pulls his hands back while catching a ball, increasing $\Delta t$, so the force on his hands is reduced.
  • Vehicles have crumple zones and we use airbags so that the stopping time increases and the force on passengers falls.
  • A high jumper lands on a thick foam mattress so the longer stopping time lowers the impact force.
  • Athletes are advised to bend their knees when landing to prolong contact and reduce force.
Newton's Second Law: same force gives larger acceleration to smaller massNewton's Second Law: F = maSmall mass (2 kg)2 kgF = 10 Na = 5 m/s²Large mass (5 kg)5 kgF = 10 Na = 2 m/s²Momentum p = mvSecond Law: F = change in momentum / time = maSame force: smaller mass gets bigger acceleration1 newton = 1 kg x 1 m/s²
1
Worked Example
Calculate the momentum of a $5\ \text{kg}$ ball moving at $4\ \text{m/s}$.
Solution
  1. Use $p = mv$.
  2. Substitute $m = 5\ \text{kg}$ and $v = 4\ \text{m/s}$.
  3. $p = 5 \times 4 = 20\ \text{kg m/s}$.

Answer: $p = 20\ \text{kg m/s}$ in the direction of motion.

2
Worked Example
A force acts on a $2\ \text{kg}$ body and gives it an acceleration of $3\ \text{m/s}^2$. Find the force.
Solution
  1. Use Newton's Second Law $F = ma$.
  2. Substitute $m = 2\ \text{kg}$ and $a = 3\ \text{m/s}^2$.
  3. $F = 2 \times 3 = 6\ \text{N}$.

Answer: $F = 6\ \text{N}$.

3
Worked Example
A car of mass $1000\ \text{kg}$ speeds up from $5\ \text{m/s}$ to $15\ \text{m/s}$ in $5\ \text{s}$. Find the force applied by the engine.
Solution
  1. Acceleration $a = \frac{v - u}{t} = \frac{15 - 5}{5} = 2\ \text{m/s}^2$.
  2. Apply $F = ma$.
  3. $F = 1000 \times 2 = 2000\ \text{N}$.

Answer: $F = 2000\ \text{N}$ in the direction of motion.

4
Worked Example
A bullet of mass $0.02\ \text{kg}$ moving at $100\ \text{m/s}$ is stopped in $0.05\ \text{s}$ by a wooden block. Find the average force exerted by the block on the bullet.
Solution
  1. Initial momentum $= mu = 0.02 \times 100 = 2\ \text{kg m/s}$; final momentum $= 0$.
  2. Change in momentum $\Delta p = 0 - 2 = -2\ \text{kg m/s}$.
  3. $F = \frac{\Delta p}{\Delta t} = \frac{-2}{0.05} = -40\ \text{N}$.

Answer: The block exerts an average force of $40\ \text{N}$ opposing the bullet's motion.

5
Worked Example
What force is needed to give a $50\ \text{kg}$ trolley an acceleration of $0.5\ \text{m/s}^2$?
Solution
  1. Use $F = ma$.
  2. Substitute $m = 50\ \text{kg}$ and $a = 0.5\ \text{m/s}^2$.
  3. $F = 50 \times 0.5 = 25\ \text{N}$.

Answer: $F = 25\ \text{N}$.

6
Worked Example
A cricketer lowers his hands while catching a fast ball. Using $F = \frac{\Delta p}{\Delta t}$, explain how this protects his hands.
Solution
  1. The ball's change in momentum $\Delta p$ (from fast to zero) is fixed by its mass and speed.
  2. By pulling his hands back, the cricketer increases the stopping time $\Delta t$.
  3. Since $F = \frac{\Delta p}{\Delta t}$, a larger $\Delta t$ gives a smaller force $F$.

Answer: Increasing the contact time reduces the force on his hands, so they are not hurt.

Key Points

  • Momentum $p = mv$ is a vector; its SI unit is $\text{kg m/s}$.
  • Newton's Second Law: force equals the rate of change of momentum, $F = \frac{\Delta p}{\Delta t}$.
  • For constant mass this gives $F = ma$, so force = mass × acceleration.
  • One newton gives a $1\ \text{kg}$ mass an acceleration of $1\ \text{m/s}^2$: $1\ \text{N} = 1\ \text{kg m/s}^2$.
  • Increasing the time of contact reduces the force (catching a ball, airbags, foam landing mats).
Tap an option to check your answer0 / 4
Q1.The momentum of a body is given by:
Explanation: Momentum is mass times velocity, $p = mv$.
Q2.The SI unit of momentum is:
Explanation: Momentum is measured in $\text{kg m/s}$.
Q3.According to Newton's Second Law, force equals:
Explanation: $F = \frac{\Delta p}{\Delta t}$, the rate of change of momentum.
Q4.A $4\ \text{kg}$ body has an acceleration of $5\ \text{m/s}^2$. The force on it is:
Explanation: $F = ma = 4 \times 5 = 20\ \text{N}$.