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:
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.
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?
Solution:
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?
Solution:
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.