The resultant force causes a change in
speed or direction in the direction of the resultant force.

**The resultant force causes a change in
momentum in the direction of the resultant force.**

A way of stating Newton's 2nd law is to
say the change in momentum is proportional to the size of the resultant
force and the time interval during which the force is acting on the object.

This statement is justified by
following the three mathematical steps below.

A resultant force on an object will cause it
to accelerate or decelerate.

force = mass x acceleration

**
F = ma** (the
mathematical expression of Newton's 2nd Law of Motion)

NOW acceleration is the change in velocity in
a specific change in time.

**
a =
∆v / ∆t**
(the usual formula for acceleration)

**a** = acceleration in m/s^{2},
**v** in m/s, time **t** in seconds s

If you combine these two equations you get

force = mass x change in velocity / time

**
F = m **
**
∆v / ∆t = m(v-u) / ∆t**

(where v and u are the final and initial velocities)

but, for a given
object of mass m, m ∆v
**= ****∆mv = ∆p = change in momentum**

so can write Newton's
2nd law equation as

**F =
****∆mv / ∆t = ∆p / ∆t
** (force is equal to the rate of change of momentum)

**
These equations express force as the
rate of change of momentum**

and
**
∆p = ∆mv = F x ∆t
**

(remember ∆t is the time over which the resultant force acts)

**F x ∆t** is sometimes called the
**impulse**.

**This means that a force applied to an
object over any period of time must change that object's velocity.**

From the point of view of solving calculation
problems you need to be pretty familiar with all this maths!

In calculations you can then use Newton’s 2nd
law of motion (F = ma) as follows:

force (newton, N) = change in
momentum (kilogram metre per second, kg m/s) / time (second, s)

**F = ∆mv / ∆t =
(mv - mu) / ∆t**, where = mass of object kg, u = initial velocity
m/s, v = final velocity m/s

this means the **rate of change of
momentum is a directly related to the resultant force or force applied**,

so you can say **the force equals the
rate of change of momentum**,

the equation can be written as** F = mΔv
/ Δt ** since m is constant for a given object

which can be expressed as ...

change of momentum (kg m/s) **=**
resultant force (N) **×** time for which it acts (s)

**
∆p = ∆mv = F x
**
**
∆t**

(the
product of the resultant force x time is sometimes called the **impulse**)

(if the force is
varying, then the average force is used in the calculation, but I don't
think this is needed for GCSE)

For any moving object the faster the
change in momentum (Δmv) occurs, the greater the force (F) involved,
and the shorter the time taken (∆t).

You should be able to see this from
the equation ∆mv = F x
∆t

for a given
mv, increase in F means a decrease in ∆t,

for a given
mv, decrease in F means an increase in ∆t,

*This has serious implications for e.g.
car crashes.*

The faster the crash happens the
bigger the force on the car and its occupants.

This sudden change momentum change in
such a short time results in a great impact force.

The smaller Δt for a given
momentum change, the greater the force involved and the greater the
chance of injury.

If you can make the rapid
deceleration occur over a longer time (increasing Δt) you reduce
the resultant force and decrease the chance of serious injury.

That is what crumple zones are for in
the design of modern cars.

In terms of **F = ma**, you are
trying to **reduce a** and **so decrease F**.

**
impulse = force x time**

(time x the
force is applied in the direction of action in units of **Ns**)

Since force = mass x acceleration (F
= ma),

and acceleration **
a = ****
Δv/Δt, **then ...

impulse = mass x acceleration x time, and so ...

**
impulse = mass x change in momentum =
mΔv**