2
Momentum and Impulse
2.1
Momentum
Momentum of a body is the product of its mass and velocity. It is a vector quantity and its
units are kgms −1 or newton second (Ns). The higher the velocity of the body, the more the
momentum.
Examples
1. If a particle of mass 2 kg moves velocity of 5 ms−1 along a straight road. Such a particle
has a momentum of
= 2 ∗ 5 Ns
= 10 Ns
Trial Questions
1. Determine the momentum of a
(a) 60-kg halfback moving eastward at 9 m/s.
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(c) 40-kg freshman moving southward at 2 m/s.
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(b) 1000-kg car moving northward at 20 m/s.
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2. A car possesses 20 000 units of momentum. What would be the car’s new momentum if
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(a) its velocity was doubled.
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(b) its velocity was tripled.
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(c) its mass was doubled (by adding more passengers and a greater load)
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(d) both its velocity was doubled and its mass was doubled.
2.2
Changes of Momentum
If initially (at t = 0) the body is moving with velocity u and after time t, the body attains a
velocity of v, then;
Change in Momentum = Final momentum - Initial momentum
Change in Momentum = mv − mu
or Change in Momentum = m(v − u)
(16)
This change in momentum is the change impulse.
Examples
1. For example, suppose a body of mass 2 kg changes its speed from 6 m/s to 15 m/s in the
same direction;
Then its change in Momentum = Final momentum - Initial momentum
Change in Momentum = 2 ∗ 15 − 2 ∗ 6
or Change in Momentum = 2(15 − 6) = 18 Ns
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Initial
3 m/s
5 m/s
Final
2 kg
2 kg
2. If a body of mass 2 kg changes its speed from 5 m/s to 3 m/s in the opposite direction.
Initial
Final
Change in
∴ Change in
momentum
momentum
momentum
momentum
2 × 5 = 10 N s
2 × −3 = −6 N s
−6 − 10 = −16 N s
16 N s
=
=
=
=
(17)
2.3
Impulse
Impulse is the change in momentum. i.e,
Impulse = Change in Momentum
Impulse = mv − mu = m(v − u)
(18)
From Newton’s first equation of motion, v = u + at or v − u = at, then;
(19)
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Impulse = m(v − u)
Impulse = m(at) = mat
But from Newton’s second law of motion, ma=F. ⇔ Impulse = F t
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Therefore, impulse can also be defined as the product of Force and the time in which it acts.
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1. A body of mass 2 kg is initially at rest on a smooth horizontal surface. A horizontal force
of 6 N acts on the vbody for 3 seconds. Find
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(a) the magnitude of the impulse given to the body,
(b) the magnitude of the final momentum of the body,
(c) the final speed of the body.
2. A body of mass 10 kg is initially moving with constant velocity (2i-7j) m/s. The body
experiences a force of (5i+10j) N for 4 seconds. Find the final velocity of the body in
vector form and hence obtain its final speed.
2.4
Conservation of Linear momentum
The law of linear momentum states that when two or more bodies collide, the total momentum
before collision is equal to the total momentum after collision provided no external forces act
on the system.
Consider two bodies A and B of masses mA and mB moving initially with velocities uA and
uB , respectively. Suppose these bodies collide and their final velocities are vA and vB , then
from the law of conservation of linear momentum;
Initial momentum = Final momentum
mA uA + mB uB = mA vA + mB vB
(20)
Note that if the collision is elastic, there is no loss of kinetic energy. Implying that for an
elastic collision, the total kinetic energy of the system before collision is equal to the total
kinetic energy after collision.
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If the Collision is inelastic, then there is loss of kinetic kinetic energy. This energy could be
lost to sound, heat or even light. Then the lost kinetic energy k.e can be calculated as follows;
Loss = Initail
k.e - Final k.e 1
1
1
1
2
2
2
2
mA uA + mB uB −
mA vA + mB vB
Loss =
2
2
2
2
(21)
Note: The kinetic energy of a body of mass m moving with velocity u is; 21 mu2
1. A particle A of mass 300 g lies at rest on a smooth horizontal surface. A second particle
B of mass 200 g is projected along the surface with speed 6 m/s and collide directly with
A. If the collision reduces B to rest, find
(a) the speed with which A moves after collision,
(b) the initial kinetic energy of the system,
(c) the final kinetic energy of the system,
(d) the loss in the kinetic energy of the system during the collision.
Solution: (a) From the conservation of linear momentum,
After collision
Before Collision
300 g
200 g
rest
v ms−1
300 g
= Final momentum
= mA vA + mB vB
300
200
=
×v+
×0
1000
1000
= 4 ms−2
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Initial momentum
mA uA + mB uB
200
300
×0+
×6
1000
1000
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6 ms−1
200 g
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B
(22)
(b) The initial kinetic energy is determined using the formula k.e = 12 mu2 ;
1
1
mA u2A + mB u2B
2
2
1
300
1
200
Initail k.e =
×
× 02 + ×
× 62
2 1000
2 1000
Initail k.e = 3.6 J
Initail k.e =
(c) The final kinetic energy is determined using the formula k.e = 21 mv2 ;
1
1
2
mA vA
+ mB vB2
2
2
1
300
1
200
Initail k.e =
×
× 42 + ×
× 02
2 1000
2 1000
Initail k.e = 2.4 J
Initail k.e =
(d) The loss in k.e = Initial K.e - Final k.e
Loss = 3.6 - 2.4
Loss = 1.2 Joules
2. A shell of mass 5 kg is fired from a gun of mass 2000 kg. The shell leaves the gun with a
speed of 400 m/s. Find the initial speed of recoil of the gun and the gain in the kinetic
energy of the system.
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Solution:
Let the speed of recoil of the gun be V
From the law of conservation of linear momentum,
Initial Momentum = Final momentum
0 = momentum of the gun + Momentum of recoil of the gun
0 = 5 × 400 + (−V) × 2000
5 × 400
V =
= 1 ms−1
2000
Trial Questions
1. A ball of mass 0.2 kg is moving at 4 m/s when it collides directly with a stationary ball
B of mass 0.1 kg. After collision, A moves with a velocity 2.5 m/s in the same direction
as before. Calculate
(a) the speed of B after the collision,
(b) the total loss of kinetic energy, in J, due to the collision.
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2. A bullet is fired horizontally with speed of 600 m/s into a block of wood of mass 0.245 kg,
resting on a smooth horizontal plane, and becomes embedded in the block. Given that
the block begins to move with speed of 12 m/s , find , in kg , the mass of the bullet.
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3. Two trucks P and Q, of mass 150 g and 250 g respectively, are free to move on a straight
horizontal track of a model railway. Initially Q is at rest and P is moving towards Q with
a velocity of 40 cm/s. Immediately after the impact Q has a velocity of 8 cm/s relative
to P. Calculate
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(a) the velocities of each truck after collision,
(b) the impulse imparted to each truck by the impact.
4. A particle of mass m moving with speed v strikes a particle of mass 3m at rest and
coalesces with it. Express the final kinetic energy as a fraction of the original kinetic
energy.
5. Two identical railway trucks are traveling in the same direction along the same straight
piece of track with constant speeds of 6 m/s and 2 m/s. The faster truck catches up with
the other one and on collision, the two trucks automatically couple together. Find the
common speed of the trucks after collision.
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