6. Elastic and nonelastic collisions, conservation of momentum  concepts and calculations and Newton's 2nd law of motion
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This page will help answer questions such as ...
What is the momentum of an object? What is an elastic collision? What is an inelastic collision? How do we solve problems and do calculations involving
momentum?
*
Subindex for this page
(a)
What is the momentum of an object and how do we calculate it?
(b)
The Law of Conservation of Momentum
(c)
Other examples of momentum effects
(d)
Problem solving questions
involving more complex momentum calculations
(e)
Change in Momentum and the consequences of Newton's Second Law of Motion
(f)
More advanced calculations involving momentum and
Newton's 2nd/3rd Laws
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(a)
What is the momentum of an object and how do we calculate it?
All moving objects have what we call
momentum. (sometimes denoted by a lower case p)
The faster an object moves, the bigger
its momentum.
The bigger the mass of a moving object,
the bigger its moment.
The momentum of an object is directly
proportional to its velocity AND its mass.
Momentum (p) is a vector quantity,
it has both size and direction
Therefore the equation for momentum is
...
momentum (kilogram metre per second) = mass (kilogram) × velocity (metre per second)
momentum p (kg m/s) = mass
m (kg) x velocity
v (m/s)
p = m × v
Example of simple calculations using the formula for
momentum
Q1.1 A 120 kg sprinter is travelling at 9.0 m/s, what is the
momentum of the runner?
momentum p = m × v
p = 120 x 9 =
360 kg m/s
Q1.2 If a 20 g bullet has a momentum of 8 kg m/s, calculate its
velocity.
p = m × v, v = p/m
(don't forget to change the g to kg!)
velocity = 8/(20/1000) =
400 m/s
Q1. 3 What mass must an object have if it has a momentum of 1.5 x 10^{6}
kg m/s and a velocity of 30 m/s?
p = m × v, m = p/v
mass = 1.5 x 10^{6}/30 =
5.0 x 10^{4}
kg
Q1.4 What is the velocity of a 1500 kg car travelling north with 30
000 kg m/s of momentum?
p = m x v, v = p / m
v = 30 000 / 1500 =
20 m/s in a northerly direction
TOP OF PAGE
and subindex
(b)
The Law of Conservation of Momentum
Here we will consider collisions between
two objects in a closed system.
Here a closed system here means no other
external forces affect the situation e.g. a collision between two objects 
the event.
If an external force like friction is
involved, total momentum cannot be conserved.
The total momentum of an event in a
closed system is the same before and after the event (e.g. a collision between
two objects).
This is called the 'Law of
Conservation of Momentum' and you can use it do lots of calculations!
e.g. total momentum of two colliding
objects = total moment of objects after collision
e.g. for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision,
(you are assuming there is no
change in mass, no bits have flown off!)
and if two objects stick together'
after the collision then: p_{1} + p_{2} = p_{3}
substituting m and v for the mass
and velocity gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{3}v_{3} (where m_{3} = m_{1} + m_{2})
where v_{1} and v_{2} are the initial
velocities and v_{3} and m_{3} (m_{3} = m_{1}
+ m_{2}) are the final velocity and mass after the collision.
In other words the large object
formed by collision has the momentum equal to the two momentums of
the colliding objects added together.
Momentum is conserved for both elastic and
inelastic collisions.
For a perfect elastic collision,
no kinetic energy is lost  kinetic energy conserved.
In an elastic collision, the
total energy in the kinetic energy stores of the colliding objects
is the same as before and after the collision.
You will not have to solve
problems for elastic collisions  the maths is too difficult for
GCSE physics, with two sets of equations, for momentum (mv) and
kinetic energy (E = ½mv^{2}), to solve e.g. for the
resultant velocities!
For an inelastic collision,
kinetic energy is not conserved  kinetic energy is lost in some form
e.g. heat or sound.
In an inelastic collision, some
of the moving objects kinetic energy stores are lost and transferred
to other energy stores of the objects themselves or the environment.
This is because the atoms are
bashed together increasing their potential energy store (compressed
for a fraction of a second). They 'relax'
to their normal state by losing the energy as heat (thermal
energy) or sound.
For inelastic collisions you can
solve a variety of problems using the principle of 'conservation of
momentum'.
TOP OF PAGE
and subindex
(c)
Other examples of momentum effects
The above examples involve collision but
there are all sorts of momentum situations e.g.
When a gun is fired the bullet goes in
the forward direction (positive momentum), but the gun recoils in the complete
opposite direction (negative momentum).
From Newton's 3rd law, you get an equal
and opposite force reaction.
The bullet goes one way and the gun the
other way from the force of the explosion.
numerically, ignoring +/ signs: m_{gun}v_{gun}
= m_{bullet}v_{bullet} (see Q2.2
below)
Note that the combined momentum = zero,
which it is at the start and immediately after the gun is fired.
Before a rocket is fired it has zero
velocity and zero momentum.
When the rocket's propellant is fired,
the rocket goes one way (positive momentum) and the exhaust gases go in the
opposite direction (negative momentum).
So, in conserving momentum, we can say ..
numerically, ignoring +/ signs: m_{rocket}v_{rocket}
= m_{exhaust gases}v_{exhaust gases}
(see Q2.2 below)
Because the rocket produces gases with
great force and velocity, the rocket must respond according to Newton's 3rd
law of motion  equal and opposite forces in dramatic action!
TOP OF PAGE
and subindex
(d) Problem solving questions involving more complex momentum
calculations
Q2.1 The diagram below shows the sequence of events when a moving
green ball collides with a stationary purple ball.
The green ball has a mass of 2.0 kg and is moving at 4.0 m/s prior
to colliding with the stationary purple ball.
After the collision both balls are moving in
a forward direction.
If the green ball is still moving forward at 1.5 m/s,
calculate
the velocity of the purple ball.
From the law of conservation of momentum
total momentum before collision = total
momentum after collision
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{1}v_{4}
(2 x 4) + (3 x 0) = (2 x 1.5) + (3 x v_{4})
8 + 0 = 3 + 3v_{4}, 3v_{4}
= 8  3 = 5, therefore V_{4} = 5/3 =
1.7 m/s (2sf)
Q2.2(a) Why does a gun recoil (move backwards) when fired?
At the start both the gun and the
bullet have zero moment.
After firing the bullet then has its
own momentum in a forward direction from the shooter.
BUT, this must be balanced by a
momentum in the opposite direction because total momentum must be
conserved.
So, on firing, the gun itself must
get its own equal but opposite momentum (remember that momentum is a
vector quantity!).
Therefore the gun's equal and
opposite momentum causes it to recoil backwards.
(b) A 10 g bullet accelerates to 400 m/s
after being fired from a gun with a mass of 2.0 kg.
(i) What is the momentum of the
bullet? (kg = g/1000)
p = m × v = (10/1000) x
400 =
4.0 kg m/s
(ii) What is the momentum of the gun?
From the law of conservation of
momentum, the gun must have a momentum equal and opposite to that of
the bullet.
Therefore gun momentum =
4.0 kg m/s
(the minus sign is important,
particularly in more complex calculations, the gun recoils in the
opposite direction to the bullet  Newton's 3rd Law in action).
(iii) What is the velocity of the
recoiling gun?
The momentum of the gun and
bullet must be equal
(from the law of conservation of
momentum)
therefore numerically: m_{gun}v_{gun}
= m_{bullet}v_{bullet}
m_{gun}v_{gun} =
4.0 = 2.0 x v_{gun}, v_{gun} = 4.0/2.0 =
2.0 m/s
(again, note the minus sign,
because the gun recoils in the opposite direction to the bullet).
(c) Why is it an advantages to the gunner
to fire with
a heavier gun?
From the equation: momentum = mass x
velocity
The bigger the mass of the gun,
for the same momentum, the lower the recoil velocity and the lesser
impact on the person firing the gun.
Q2.3
The diagram below shows the sequence of events when a moving car crashes into a
stationary car and they combine together and move forward.
The green car (1), of mass 1000 kg, crashes into
the stationary blue car (2) of mass 800 kg.
After the collision, the two cars are locked
together.
(a) Calculate the velocity of the
combined wrecked cars
immediately after the condition.
From the law of conservation of momentum
total momentum before collision = total
momentum after collision
m_{1}v_{1} + m_{2}v_{2}
= m_{3}v_{3} (m_{3} = m_{1}
+ m_{2})
(1000 x 20) + (800 x 0) = (1000 + 800) x
v_{3}
20 000 + 0 = 1800v_{3}, v_{3}
= 20 000/1800 = 11.1 m/s
(3sf)
(b) Calculate the kinetic energy of all
three objects in the diagram.
KE = ½mv^{2}
KE_{green car} = ½ x 1000 x
20^{2}, =
200 000 J
KE_{blue car} = ½ x 800 x 0^{2}
= 0 J
KE_{wreck} = ½ x 1800 x 11.1^{2},
= 110 889 = 111 000
J (to 3 sf)
(c) Calculate the difference, if any, of
the total initial kinetic energies of the cars and that of the wreck,
and comment on the results of your
calculations.
Kinetic energy is NOT conserved.
The loss of kinetic energy = 200 000
 110 889 = 89 111 = 89 100 J
(to 3 sf)
Lots of sound and thermal energy
created!
(d) Explain your answer to (c)
The loss of kinetic energy involves
several changes of energy stores.
There is a loud bang on impact, so
some energy is lost as sound.
The cars are compressed and the
increase in potential energy plus friction effects, increases the
thermal energy store of the wreck  which then cools down.
So, in the end, all the lost kinetic
energy of the cars energy stores increases the thermal energy store of
the surroundings  lost KE somewhat dissipated!
Q2.4 Imagine a 0.2 kg ball moving at 5 m/s colliding with a 0.3 kg
ball moving at 2 m/s in the same direction.
After the collision the 0.2 kg green ball stops moving and the 0.3 kg
purple ball continues moving in the same direction.
Calculate the final velocity of the of the 0.3 kg
purple ball.
First sort out all the masses and velocities
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
m_{1} = 0.2 kg, initial velocity
v_{1} = 5 m/s, final velocity v_{3} = 0 m/s
m_{2} = 0.3 kg, initial velocity
v_{2} = 2 m/s, final velocity v_{4} = ? m/s
total momentum of two of objects before
collision = total moment of objects after collision
for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity of the two objects gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
Then substitute everything in the equation,
rearrange and deduce the answer (v_{4}).
(0.2 x 5) + (0.3 x 2) = (0.2 x 0) + (0.3
x v_{4})
1.0 + 0.6 = 0 + 0.3v_{4},
0.3v_{4} = 1.6
therefore the final velocity of the 0.3
kg ball = v_{4} = 1.6/0.3 =
5.3 m/s (2 sf)
Q2.5 Imagine a 0.2 kg ball moving at 5 m/s hitting a 0.3 kg ball
moving at 2 m/s in the same direction.
After the collision both balls keep moving in the same
direction, but with different velocities.
If the 0.2 kg ball continues to move at a velocity of 1.5 m/s,
calculate the final velocity of the 0.3 kg ball.
This is an almost identical problem to Q2.4, except there is no
zero momentum term.
So, sorting out all the masses and velocities
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
m_{1} = 0.2 kg, initial velocity
v_{1} = 5 m/s, final velocity v_{3} = 1.5 m/s
m_{2} = 0.3 kg, initial velocity
v_{2} = 2 m/s, final velocity v_{4} = ? m/s
total momentum of two objects before
collision = total moment of objects after collision
for two colliding objects, where
p = momentum: p_{1} + p_{2} = p_{3} + p_{4}
substituting m and v for the mass
and velocity of the two objects gives ...
m_{1}v_{1} + m_{2}v_{2}
= m_{1}v_{3} + m_{2}v_{4}
where v_{1} and v_{2}
are the initial velocities and v_{3} and v_{4} the
velocities after the collision
Then substitute everything in the equation,
rearrange and deduce the answer (v_{4}).
(0.2 x 5) + (0.3 x 2) = (0.2 x 1.5) +
(0.3 x v_{4})
1.0 + 0.6 = 0.3 + 0.3v_{4},
0.3v_{4} = 1.6  0.3 = 1.3
therefore the final velocity of the 0.3
kg ball = v_{4} = 1.3/0.3 =
4.3 m/s (2 sf)
Note the final velocity of the 0.3 kg
ball is less than in Q2.4 because in Q2.4 it received all the momentum from
the collision.
Q2.6 Snooker players can hit the white ball 'dead centre' into a
stationary red ball. The stationary red ball then flies off at the same velocity
as the incoming white ball and the white ball is left in a stationary position
where the red ball was.
Ignoring friction and assuming the balls
have the same mass, explain this observation.
The law of conservation states the total
momentum must be the same before and after the collision.
m_{white}v_{intitial} + m_{red}v_{initial}
= m_{white}v_{final} + m_{red}v_{final}
m_{red}v_{initial}
is zero, since the red ball is initially stationary with a momentum of zero.
m_{white}v_{final}
is zero, since the white ball is now stationary, also with a momentum of
zero.
Therefore: m_{white}v_{intitial}
= m_{red}v_{final}
and since the masses are equal, m_{white}
= m_{red}, then the final velocity of the red ball must equal
the initial velocity of the white ball.
It should be noted that skilled snooker
players can play all sorts of tricks with the 'physics' of snooker and often
have a total disregard for the law of conservation of momentum!
Q2.7
Imagine a 1000 kg car travelling with a velocity of 10 m/s crashing headon into
a 500 kg car travelling in the opposite direction with a velocity of 15 m/s.
Assume they 'crunch' together to form a single object  the 'wreck'!
(a) Calculate the momentum of both cars.
(i) green car momentum = 1000 x 10 =
+10 000 kg m/s (to the right)
(ii) blue car momentum = 500 x 15 =
7 500 kg m/s (to the left)
(you must assign a positive and
negative signs, vector quantity, opposite directions)
(b) Calculate the momentum p of
the 'wreck'
From the law of conservation of
momentum
p_{wreck} = p_{green car}
+ p_{blue car}
p_{wreck} = +10 000 + 7 500
=
+2
500 kg m/s
(c) What is the immediate velocity of the
wreck after the impact and in what direction? (in terms of diagram)
total mass = 1000 + 500 = 1500 kg
p_{wreck} = m_{wreck}
x v_{wreck}
2500 = 1500v, v = 2500 / 1500 = 1.6r
=
1.67 m/s to the right (in terms of diagram and to 3 sf
too!)
Note: Although the speed of the blue
car is greater than the green, it has a greater momentum, so the
resultant movement of the 'wreck' is to the right.
(d) Is kinetic energy conserved in this
collision?
(i) You need to calculate the kinetic
energies of all the objects concerned
KE = ½mv^{2}
KE_{green car} = ½ x 1000
x 10^{2}, =
50
000 J
KE_{blue car} = ½ x 500 x
15^{2} = 56
250 J
KE_{wreck} = ½ x 1500 x
1.67^{2}, = 110 889 =
2 092 J
(to 3 sf)
(ii) Kinetic energy is NOT conserved.
In fact, most of the kinetic
energy of the colliding cars is lost on impact.
Reminder of diagram
A reminder of the data and the final
velocity was 1.67 m/s to the right!
(e) If the impact time is 0.10 seconds
calculate the decelerations of the two cars.
(velocities are vector quantities,
watch the signs, + to right,  to the left).
a = ∆v / ∆t = (v 
u) / ∆t (u = initial velocity, v = final velocity)
a_{green}
car = (1.6r  10) / 0.1 =
83.3 m/s^{2} (3 sf, r meaning
recurring)
a_{blue}
car = (1.6r  15) = 16.6r / 0.1 = 166.6r =
167 m/s^{2}
(3 sf, r meaning recurring)
(f) Calculate the
impact force experienced by each car.
F = ma
F_{green}
car = 1000 x 83.3r = 83 333 =
83 300 N (3 sf)
F_{blue}
car = 500 x 166.7 = 83 333 =
83 300 N (3 sf)
(g) Comment on your answer to (f) in
terms of Newton's 3rd Law of motion.
The two forces are equal numerically,
but act in opposite directions.
Newton's Third Law of motion states that
when two objects interact, the forces they exert on each other are equal in
numerical value and act in opposite directions and in this case they both normal
contact forces.
Q2.8
A momentum question involving Newton's 3rd Law, conservation of momentum and
conservation of kinetic energy.
A 0.20 kg green ball (m1) moving at 5.0
m/s collides in line with a 0.40 kg purple ball (m2) moving at 2.0 m/s.
Also, from high speed photography, the
collision impact time was found to be 0.05 seconds.
After the collision, with both moving in
the same direction, the 0.20 kg green ball continues moving forwards at a
slower speed of 2.0 m/s and the purple ball also continues moving forwards
at an increased speed of 3.50 m/s.
(a) Using Newton's 2nd Law, calculate the
force involved in causing the changes in velocity.
F = ma = m∆v/∆t =
m(vu) ÷ 0.05, where F = force (N), m = mass
(kg), v and u = final and initial velocities.
For the green ball: F = 0.2 x (2  5) ÷
0.05 = 12 N
For the purple ball: F = 0.4 x (3.5 
2) ÷ 0.05 =
12 N
(b) Do the results of your calculations
in (a) comply with Newton's 3rd Law of Motion?
Yes they do. See notes on
Newton's 3rd Law
Newton's Third Law of motion states that
when two objects interact, the forces they exert on each other are equal in
numerical value, act in opposite directions and are of the same type.
The impact forces are the same size and act in opposite directions.
The impact forces act on two different objects
with same type of normal contact force.
Data
reminder!
(c) Does the experiment comply with (i)
the concept of a 'closed' system AND (ii) do the observations comply with
the law of conservation of momentum?
(i)
The collision does take place
in a closed system  the experiment is a
physical system that does not allow transfer of matter in or out of the
system, and is so fast there is no time for energy to leave or enter the
system.
(ii) momentum p = mv, where p
in kg m/s, m in kg, v in m/s.
Momentums before collision:
for green ball p = 0.2 x 5 =
1.0 kg m/s
for purple ball p = 0.4 x 2 =
0.8 kg m/s
total momentum before
collision = 1.0 + 0.8 =
1.8 k m/s
Momentums after collision:
for green ball p = 0.2 x 2 =
0.4 kg m/s
for purple ball p = 0.4 x 3.5
= 1.4 kg m/s
total momentum after
collision = 0.4 + 1.4 =
1.8 k m/s
The total momentum is the same as
before and after the collision, so the
law of conservation of momentum is obeyed.
(d) Show by calculation whether kinetic
energy is conserved or otherwise? Comment on your results.
KE = ½mv^{2}, where KE in J,
m = mass of object in kg and velocity in m/s.
Kinetic energies before collision:
green ball KE = ½ x 0.2 x 5^{2}
= 2.5 J
purple ball KE = ½ x 0.4 x 2^{2}
= 0.8 J
total kinetic energy before
collision = 2.5 + 0.8 =
3.3 J
Kinetic energies after collision:
green ball KE = ½ x 0.2 x 2^{2}
= 0.4 J
purple ball KE = ½ x 0.4 x 3.5^{2}
= 2.45 J
total kinetic energy after
collision = 0.4 + 2.45 =
2.85 J (2.9 2sf)
You can see from the calculations
that
kinetic energy is not conserved in this collision of two objects.
This is an
inelastic collision.
The atoms of the object are
compressed and kinetic energy is converted into potential energy,
thermal energy and sound.
TOP OF PAGE
and subindex
(e)
Change in Momentum and the consequences of Newton's Second Law of Motion
When a resultant force acts on an object
for any length of time it will cause a change in momentum by changing the
velocity.
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(vu) / ∆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)
Consequences of the force
equalling the rate of change of momentum
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.
(For more details on this issue see
5. Stopping distances,
impact forces  example
calculations
TOP OF PAGE
and subindex
(f) More advanced calculations involving momentum and
Newton's 2nd/3rd Laws
Q3.1
What explosive force is required to accelerate a 20 g bullet in a gun
from 0 m/s to 300 m/s in 0.15 seconds.
20 g = 20/1000 = 0.02 kg, u = 0 m/s, v =
300 m/s, ∆t = 0.15 s
F = ∆mv / ∆t = (mv  mu) /
∆t
F = {(0.02 x 300)  (0.02 x 0)} / 0.15
F = (0.02 x 300) / 0.15 = 6.0 / 0.15 =
40 N
It is important that you can solve
these sorts of problems in different ways, depending on what information
you get and how you are expected to deduce things from it e.g.
Using the same information you
can solve this problem using F = ma directly.
a = 300 / 0.15 = 2000 m/s^{2}, m =
0.02 kg, F = ma, F = 0.02 x 2000 = 40 N
This is the approach I took on my
stopping distances, impact forces  example calculations
Q3.2 What force must a tennis player generate with a tennis racket
exert on a 60 g tennis ball to accelerate it from rest to a speed of 50 m/s in
10 milliseconds (10 ms).
mass of tennis ball =
60 g = 60/1000 = 0.06 kg, time force applied = 10 ms = 10/1000 = 0.01 s
∆p = ∆mv = F x
∆t (this comes from Newton's 2nd Law equation F = m∆v/∆t)
mv changes from zero
to 0.06 x 50 = 3.0 kg m/s
therefore: 3.0 = F x
0.01, F = 3.0 / 0.01 =
300 N
Q3.3 An object experiences a change in momentum of 30 kg m/s.
If the change takes place in 5
seconds, what force on the object produced this change in momentum?
F = ma = m∆/∆t =
∆mv / ∆t = 30 / 5 =
6 N
Q3.4 A force of 500 N is applied to an object to change its momentum
by 2000 kg m/s.
For how long was the force applied?
F = ma = m∆v/∆t =
∆mv/∆t
∆t = ∆mv / F =
2000 / 500 = 4 seconds
Q3.5 A 0.80 g lead pellet
fired from an airgun experiences a force of 4.8 N in a time interval of 25
milliseconds  the time it takes for the pellet to travel down the barrel.
(a) Calculate the
speed at which the pellet leaves the gun.
F = ma =
m∆v / ∆t
= m(vu) / ∆t
m = 0.8 / 1000 =
0.0008 kg,
∆t = 25 / 1000 = 0.025 s (25
milliseconds), v = ?, u = 0 m/s
F = mv /
∆t
v = ∆t x F / m = 0.025 x 4.8
/ 0.0008 =
150
m/s
(b) If the airgun
recoils at a speed of 0.05 m/s, what is the mass of the airgun?
You can solve this
part of the problem two ways
(i) From the law
of conservation of momentum
momentum of
airgun = momentum of pellet
m x 0.05 =
0.0008 x 150
m = 0.0008 x
150 / 0.05 = 2.4
kg
(ii) From Newton's
3rd Law, which tells you that the force on the gun is equal in size to
the force on the lead pellet.
F = ma, m = F
/ a
m =? kg, F =
4.8 N, a =
∆v / ∆t =
0.05 / 0.025 = 2.0 m/s
m = 4.8 / 2.0
=
2.4 kg
For lots more questions like these, see section
5(h). calculations including F = ma
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