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Physics Lesson Notes on Linear Momentum
Linear momentum
Newton’s first law: A body will continue in its state of rest, if it is motion, will continue to
move with uniform speed in a straight line unless it is acted upon by a force
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change in momentum
Force ∞
time
Newton’s third law: to every action, there is an equal and opposite reaction
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The
S
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Momentum = mass (m) x velocity (v)
From Newton’s 2nd law, we can derive equation for force and impulse
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If an object of mass m, with initial velocity u, is acted upon by a force, F,
it will attain a final velocity, v in time t
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It is also equal to the change in momentum produced by the force
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momentum before collision = momentum before collision
m1u1 +m2 u2 = m1v1 + m2 v2

Exercises
(1)A net force of 15N acts upon a body of mass 3kg for 5s, calculate the change in
the speed of the body
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time= 5s, u= 0, v =?
F=

𝑚𝑣−𝑚𝑢

15=

𝑡
3𝑥 𝑣−3𝑥 0
5

3𝑣

15 = 5
15 x 5 = 3v
75= 3v
3v = 75
75
V= 3
V = 25ms-1
(2) A stationary object of mass 4kg is set in motion by a net force of 50N
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Solution
Force = 50N, mass = 4 kg, v = 5ms-1 t = ? u= 0
𝑚𝑣−𝑚𝑢
F= 𝑡
4x 5−4 x 0

50=
t
50t = 20
20
T = 50
2

T=5
T =0
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Calculate the magnitude of the force applied
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What is its
speed when it has travelled a distance of 8m?
Solution
F= 9N, v =? m = 4kg, d = 8m
F =ma
9=4xa
4a = 9
9
a=4
9

the acceleration= 4
v2 = u2 + 2as
u =0, it accelerated from rest
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What is the velocity of the block at t = 5 seconds?
Solution
mv−mu
F=
t

F= 16N, M= 40kg, t =5seconds, u=0, v=?
40 𝑥 𝑣+40 𝑥 0
16 =
5
40𝑣

16 = 5
16 = 8v
8v = 16
V = 2ms-1
(6)
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Calculate the force required to bring it to rest
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If the particle moves in the same direction, calculate the magnitude of the acceleration
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The resultant force = 122 + 52
= 144 + 25
= √169
= 13N
F =ma, m = 5kg, a =?
13 = 5 x a
5a = 13
13
a=
5
acceleration = 2
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5kg moving at lOm/s collides with another ball of equal mass at
rest
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Solution
Initial momentum = final momentum
m1u1 +m2 u2 = (m1 + m2)v
u2 = 0, the ball at rest before the collision
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5kg, m2 = 0
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15 x 10 + 0
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15+ 0
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5 =0
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3v = 1
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5
V =0
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0ms-2 collides perfectly with a stationary trolley of the same mass on the same platform
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Solution
The formula for the collision = m1u1 +m2 u2 = (m1 + m2)v
M1= 4kg
u1 = 1
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0 + 4 x 0 = (4 +4)v
4 = 8v
8v = 4
V = 4/8
V = 0
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5
= 8 x 0
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05kg is fired with a velocity of 200ms-1 into a lead block
of mass 0
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Given that the lead block can move freely, calculate the final K
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m1u1 +m2 u2 = (m1 + m2) v
m1=0
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95kg, u2 = 0, v =?
= 0
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95 x 0 = (0
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95) v
10 = 1v
V = 10 ms-1
K
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What is the recoil velocity of the gun?
Solution
Mass of bullet, 100g = 0
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1 x 250 = 20 x V
25 = 20V
20V = 25
V = 25/20
V =1
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25m/s due North
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W — F — mg
If a lift is accelerating upward (ascending), the weight of the body is
F = mg + ma = m(g + a)
If a lift is accelerating downwards (descending) with acceleration a, less than
g, the weight of the body is
F = mg — ma = m(g — a)
If a lift is moving downward with acceleration a greater than g, the weight of
the body is
F = ma — mg
If a lift falls freely, when, a = g then the weight of the object is
F = mg — mg = ON

(12) A spring balance, which is suspended from the roof of a lift, carries a mass of 1kg at its free
end
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5ms2, determine the reading on the spring balance
[g=10m/s2]
Mass, m = 1kg, g=10m/s, acceleration of lift, a = 2
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5) = 10 x 12
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5N
(13) An elevator of mass 4800kg is supported by a cable which can safely withstand a maximum
tension of 60 000N
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Solution
Force or tension, F = 60 000N, mass, m = 4800kg, g=10m/s2,
upward acceleration, a = ?
F = mg + ma
60
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000 = 48,000 + 4800a
4800a = 60,000-- 48,000
4800a = 12000
12000
A = 4800
a = 2
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0kg hits a smooth vertical wall normally with a speed of 2ms-1
rebound with the same speed
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Solution
Mass of ball, m = 5
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15kg is kicked against a rigid vertical wall with a horizontal velocity of
50ms-1
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Solution
Mass, m = 0
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15 x [50-(-30)] = (50 + 30) x 0
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15 x 80 =
12Ns



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