Force, Momentum and Impulse

This topic is part of the HSC Physics course under the section Momentum, Energy and Simple Systems.

HSC Physics Syllabus

  • evaluate the effects of forces involved in collisions and other interactions, and analyse quantitatively the interactions using the concept of impulse `Δp=F_{net}Δt` 

Force, Momentum and Impulse 

Force and Momentum

Force, by Newton's second law, is given by:

$$F = m \times a$$

Acceleration is the rate of change of velocity:

$$a = \frac{\Delta v}{\Delta t}$$

$$F = m \times \frac{\Delta v}{\Delta t}$$

Momentum is mass times velocity:

$$p = m \times v$$

The rate of change of momentum:

$$\frac{\Delta p}{\Delta t} = m \times \frac{\Delta v}{\Delta t}$$

Therefore, force is the rate of change of momentum:

$$F = \frac{\Delta p}{\Delta t}$$

Momentum vs Time Graph

The gradient of a line on a momentum vs time graph represents the rate of change of momentum. Therefore, the gradient in this case represents force.

 momentum vs time graph


For example, the straight line on the graph above suggests that a constant force acts on the mass. 

What is Impulse?

Impulse (I): Impulse is the change in an object's momentum, and it's equal to the product of force (F) and the time (t) during which it acts. Impulse is also a vector quantity, and its direction is the same as the force.

The equation for impulse is:

$$I = \Delta p$$

From the equation for force in terms of p and t, we can derive another equation for impulse:

$$I = F\Delta t$$

The unit for impulse is the same as momentum (kg m s-1) or the product of units of force and time (N s)

Example 1

A 100 kg mass at rest is acted upon by a 500 N net force to the right as shown for 2.5 seconds.

What is the impulse of the force and the final velocity of the mass?


First, we can calculate impulse:

$$I = F \times t$$

$$I = (500)(2.5) = 1250 \, \text{N s to the right}$$

Consider impulse as the change in momentum:

$$I = \Delta p$$

Since the mass was at rest, its initial momentum was zero:

$$I = mv$$

$$1250 = (100)v$$

$$v = 12.5 \, m \, s^{-1}$$

Example 2

An 800 kg car moving at 30 m s–1 collides with a wall. The collision causes the car to come to an abrupt stop in 0.25 s.

What magnitude of force does the wall exert on the car during the collision?


Consider the change in momentum (impulse) in terms of force and time:

$$\Delta p = F \times t$$

$$mv - mu = F \times t$$

$$(800)(0) - (800)(30) = F(0.25)$$

$$F = - 9.6 \times 10^4 \, N$$

Example 3

A 0.5 kg ball moving at 3 m s–1 on a frictionless surface collides with a vertical wall as shown. The wall exerts an average force of 120 N on the ball for a duration of 20 ms.

(a) What is the final speed of the ball?

(b) How much mechanical energy was lost during the collision?

Solution to part (a):

Consider the change in momentum as the product of force and time:

$$mv - mu = F \times t$$

Let vectors directed to the right be positive (and left to be negative). When the ball collides with the wall, the wall exerts a force on the ball to the left. 

$$(0.5)v - (0.5)(3) = (-120)(20 \times 10^{-3})$$

$$v = -1.8 \, m \, s^{-1}$$


Solution to part (b):

Mechanical energy is given by the sum of a mass' kinetic and potential energy. In this case, there is no changes in potential energy.

$$\text{initial kinetic energy} = \frac{1}{2}mu^2 = \frac{1}{2}(0.5)(3)^2$$

$$\text{initial kinetic energy} = 2.25 \, \text{Joules}$$

$$\text{final kinetic energy} = \frac{1}{2}(0.5)(-1.8)^2$$

$$\text{final kinetic energy} = 0.81 \, \text{Joules}$$ 

$$\text{mechanical energy lost} = 0.81 - 2.25 = -1.44 \, \text{Joules}$$

Therefore, 1.44 J of kinetic energy has been transformed to another form e.g. heat during the collision.