Download Momentum and Impulse

If you apply a force to an object, the effect it has depends on the
mass of the object and for how long you exert the force.
You apply a constant force, F Newtons for a time t seconds in the
direction of motion to a particle of mass m kg.
The particle accelerates at a ms-2.
The velocity of the particle increases from u ms-1 to v ms-1.
You have
v = u + at
𝐚=
𝒗−𝒖
𝑡
From Newton’s second law you know that F = ma
𝑚(𝒗 − 𝐮)
𝐅=
𝑡
Ft = mv – mu
Because force is a vector quantity, impulse is also a vector quantity
The IMPULSE of a force F applied for a time t is defined as Ft
Impulse = force x time = Ft
The SI unit of impulse is the newton second (Ns)
For a particle of mass m moving with velocity v,
Momentum = mv
The SI unit of momentum is the newton second (Ns)
In general
Impulse = change in momentum
Ft = mv – mu
Example 1
A spacecraft of mass 120kg is travelling in a straight line at a speed of 4ms-1. Its rocket is
fired for 5 seconds, exerting a force of 150N. Find the new velocity of the spacecraft if the
force was directed:
(a)
in the direction of motion
(b)
in the direction opposite to motion
(b)
(a) 150N
u = 4ms-1
150N
u = 4ms-1
Impulse = Change in Momentum
Ft = mv – mu
Impulse = Change in Momentum
Ft = mv – mu
150 x 5 = 120v – 120 x 4
v = 10.25ms-1
-150 x 5 = 120v – 120 x 4
v = -2.25ms-1
Example 2
A batsman hits a ball of mass 0.15kg at a speed of 40ms-1. The ball moves in a straight line
with constant speed until it strikes a net at right angles and is brought to rest in 0.5 seconds.
Find the magnitude of the impulse and the average force exerted by the netting on the ball.
40ms-1
F
0ms-1
Mass = 0.15kg
Momentum = mass x velocity
Momentum = mass x velocity
Momentum = 0.15 x 40
Momentum = 0.15 x 0
Momentum = 6Ns
Momentum = 0Ns
Impulse = Change in Momentum = 6Ns
Impulse = Force x time
6 = F x 0.5
F = 12N
Example 3
A particle of mass 2kg, travelling with a velocity of (3i + 5j)ms-1, is given an impulse of
(2i – 4j)Ns. Find its new velocity.
Impulse = Change in Momentum
Impulse = mv – mu
2i – 4j = 2v – 2(3i + 5j)
2i – 4j = 2v – 6i – 10j
8i + 6j = 2v
v = (4i + 3j)ms-1
Example 4
Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms-1 towards Monica. She
kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms-1 and
with an elevation of 40° to the horizontal.
Find the magnitude and direction of the impulse of Monica’s kick.
(-8cos40i + 8sin40j) ms-1
8ms-1
5i ms-1
5ms-1
Mass = 0.8kg
40°
Mass = 0.8kg
Impulse = Change in Momentum
Impulse = mv – mu
Impulse = 0.8(-8cos40i + 8sin40j) – (0.8 x 5i)
Impulse = -4.9027i + 4.1138j – 4i
Impulse = -8.9027i + 4.1138j
Example 4
Steve kicks a ball of mass 0.8kg along the ground at a velocity of 5ms-1 towards Monica. She
kicks it back towards him, but lofts it so that it leaves her foot with a speed of 8ms-1 and
with an elevation of 40° to the horizontal.
Find the magnitude and direction of the impulse of Monica’s kick.
Impulse = -8.9027i + 4.1138j
I
4.1138
θ
8.9027
I=
8.90272 + 4.11382
𝐈 = 𝟗. 𝟖𝟏𝐍𝐬
4.1138
Tan𝜃 =
8.9027
𝜽 = 𝟐𝟒. 𝟖°
Conservation of linear momentum
We will now study the motion of systems comprising of two or
more bodies, which are free to move separately
A system consists of two bodies, A and B, which have momentum
MA and MB respectively
The total momentum of the system = (MA + MB) Ns
Body A now exerts a force, F N on body B for a time t seconds.
Body B receives an impulse J = Ft Ns
By Newton’s third law, B exerts a force, -F N on A, also for time t secs
Body A receives an impulse –J Ns.
We know that
Impulse = change in momentum
The new momentum of A = (MA – J) Ns
The new momentum of B = (MB + J) Ns
The total momentum of the system = (MA – J) + (MB + J)
=
MA +
MB
i.e. the total momentum has not been changed
The forces described here are internal forces to the system.
The total momentum of a system can only be changed by the
application of an external force.
This is the principle of conservation of linear momentum.
The principle of conservation of linear momentum
The total momentum of a system in a particular direction remains
constant unless an external force is applied in that direction
In the system below:
Total momentum before collision = Total momentum after collision
mAuA + mBuB = mAvA + mBvB
Example 5
A particle of mass 4kg, travelling at a speed of 6ms-1, collides with a second particle, of mass
3kg and travelling in the opposite direction at a speed of 2ms-1. After the collision, the first
particle continues in the same direction but with its speed reduced to 1ms-1.
Find the velocity of the second particle after the collision.
Before
A
6ms-1
4kg
After
A
2ms-1
B
3kg
1ms-1
mAuA + mBuB = mAvA + mBvB
(4 x 6) + (3 x -2) = (4 x 1) + 3v
18 = 4 + 3v
v = 14/3 ms-1
B
v
Example 6
A body of mass 5kg, travelling with a speed of 6ms-1, collides with a body B, of mass 3kg,
travelling in the same direction with a speed of 4ms-1. On colliding, the two bodies
coalesce.
Find the velocity of the combined body after the collision.
Before
A
6ms-1
B
5kg
After
3kg
A+B
v
8kg
mAuA + mBuB = mA+Bv
(5 x 6) + (3 x 4) = 8v
42 = 8v
v = 5.25ms-1
4ms-1
Example 7
A railway truck of mass 4 tonnes, travelling along a straight horizontal trail at a speed of
4ms-1, meets another truck, of mass 2 tonnes, travelling in the opposite direction at a
speed of 5ms-1. They collide and become coupled together.
Find their velocity after the collision.
Before
A
4ms-1
5ms-1
4000kg
After
B
2000kg
A+B
v
6000kg
mAuA + mBuB = mA+Bv
(4000 x 4) + (2000 x -5) = 6000v
6000 = 6000v
v = 1ms-1 (in direction of A originally)
Example 8
A body of mass 4kg travelling with a velocity of (3i + 2j)ms-1 collides and coalesces with a
second body of mass 3kg travelling with velocity (i – 3j)ms-1.
Find their common velocity after impact.
mAuA + mBuB = mA+Bv
4(3i + 2j) + 3(i – 3j) = 7v
15i – j = 7v
v=
𝟏𝟓
𝒊
𝟕
𝟏
𝟕
− 𝒋 ms-1
Example 9
Two particles, A and B, of mass 3kg and 2kg respectively, lie at rest on a smooth horizontal
table. They are connected by a light inextensible string which is initially slack. B starts to
move at a speed of 8ms-1 in the direction AB. Find the common velocity of the particles
immediately after the string becomes taut, and the impulse received by each of the
particles.
Before
After
A
B
3kg
2kg
A
8ms-1
B
Momentum
= 2 x 8 = 16Ns
Before
v
3kg
2kg
Impulse = Change in momentum
(momentum after – momentum before)
(consider just one of the particles)
A = (3 x 3.2) – (3 x 0) = 9.6Ns
B = (2 x 3.2) – (2 x 8) = -9.6Ns
Impulse = 9.6Ns
Momentum
= 5v
After
Momentum
Momentum
=
Before
After
16 = 5v
v = 3.2ms-1
Example 10
A bullet of mass 50 grams is fired horizontally from a gun of mass 1kg, which is free to
move. The bullet is fired with a velocity of 250ms-1. Find the speed with which the gun
recoils.
AFTER
BEFORE
0ms-1
v
250ms-1
m = 0.05kg
m = 0.05kg
m = 1kg
Momentum
= 0Ns
Before
Momentum
Momentum
=
Before
After
0 = 12.5 – v
v = 12.5ms-1
m = 1kg
Momentum
= (250 x 0.05) + (1 x –v)
After
= 12.5 – v
Example 11
A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms-1. The gun
rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary
before the shot is fired. Find the distance the gun will move as a result of firing the shell.
AFTER
400ms-1
R
BEFORE
v
u = 0ms-1
0.6R
4g
800g
804g
Momentum = (400 x 4) + (800 x –v)
After
= 1600 – 800v
Momentum
Momentum
=
Before
After
Momentum
= 0Ns
Before
0 = 1600 – 800v
v = 2ms-1
Example 11
A gun of mass 800kg, fires a shell of mass 4kg horizontally at a speed of 400ms-1. The gun
rests on a rough horizontal surface with coefficient of friction 0.6. The gun is stationary
before the shot is fired. Find the distance the gun will move as a result of firing the shell.
R
a=?
u = 2ms-1
v = 0ms-1
0.6R
s
800g
F = ma
-0.6R = 800a
-0.6(800g) = 800a
a = -5.88ms-2
Resolving
vertically
R = 800g
u = 2ms-1 v = 0ms-1
a = -5.88ms-2 s = ?
v2 = u2 + 2as
02 = 22 + (2 x -5.88 x s)
s = 0.34m (34cm)