Transcript Slide 1

SUVAT: A ball is thrown vertically up at a speed of 11.2 ms-1,
the hand of the thrower being 1.7 m above the ground
when the ball is released. Find the time taken for the
ball to hit the ground.
Taking vertically up as the positive direction:
Consider the motion: A B
v=
u = 11.2
a = –9.8
s = –1.7
t=
A
1.7
B
Using:
s = ut + 1 at2
2
1 ( –9.8 ) t2
–1.7 = 11.2 t +
2
 4.9 t2 – 11.2 t – 1.7 = 0
49 t2 – 112 t – 17 = 0
17
The time taken is ( = 2.43 ) seconds ( 7t – 17 )( 7t + 1 ) = 0
7
Momentum : Two particles A and B of masses 3kg and 1kg
respectively are moving towards each other along the same
straight line, with speeds 7ms–1 and 1ms–1 respectively.
After impact the particles move in the same direction with
the speed of B being twice that of A. Find the magnitude of
the impulse given to A by B.
Before:
1
7
3
After:
A
1
w
B
2w
By Conservation of Momentum:
(7 × 3) + (1 × –1) = 3w + 2w
21 – 1 = 5w
w=4
Impulse on A = Change in momentum = “mv – mu”
I = 3(4) – 3(7)
= –9
=9
i.e. Mag of Impulse = 9 Ns
Summary of key points:
When two particles collide they exert equal and opposite impulses
on each other.
Impulse = Change in momentum
= Final momentum – Initial momentum
= mv – mu
The law of conservation of momentum states that when there are
no external forces acting on a system, the total momentum of the
system is constant.
For a collision:
The total momentum before a collision = the total momentum after.
This PowerPoint produced by R Collins; © ZigZag Education 2008–2010
Pulleys:
Particles A and B, of masses 2 kg and 5 kg respectively,
are connected by a light inextensible string, passing over
a smooth pulley, fixed to the top of a smooth plane,
inclined at 30 to the horizontal as shown. Particle A
moves on a line of greatest slope on the plane. The
system is released from rest with the string taut.
Find in terms of g:
i) The acceleration of the system.
ii) The tension in the string.
A
B
30°
N
T
A
a
30°
30°
2g
‘F = ma’, for B:
‘F = ma’, for A:
We need to complete
the force diagram;
T
B
a
5g
Now apply Newton’s 2nd Law:
5g – T = 5a …(1)
T – 2g sin 30 = 2a …(2)
Adding (1) + (2): 5g – 2g sin 30 = 7a
5g – g = 7a
Sub into (2):

T  g  8g 
7
4g
7
T  15g
7
a
Summary of key points:
To solve problems involving pulleys we apply Newton’s Second
Law F = ma, in the direction of the acceleration for each particle.
We always assume that the pulley is smooth. This implies that the
tension in the string is equal along its whole length.
The string is also usually considered to be inextensible. This
implies that the connected particles move with the same acceleration.
This PowerPoint produced by R Collins; © ZigZag Education 2008-2010
Moments: A uniform beam AB, of length 6 m and mass 40 kg, is at rest
on 2 supports at P and B, where AP = 1 m. A mass of 20 kg
is placed on the beam at a distance x m from B, such that the
normal reaction at P is double the reaction at B. Find the
distance x.
N
2N
A
1
P
3
2
B
x
40g
The beam is in equilibrium, so resolving:
B :
20g
3N = 60g  N = 20g
GB = 2N × 5 – 40g × 3 – 20g × x
In equilibrium, GB = 0  0 = 200g – 120g – 20g x  x = 4
Note: The position of the mass was not known, and it turns out to be to the left of
the centre of the rod.
Summary of key points:
The moment of a force about a point is defined as the product of the
magnitude of the force and the perpendicular distance of the line of
action of the force from the point.
The clockwise moment of a
force of magnitude P, about
point A, is given by:
d
A
P
A :
GA = Pd
If a rod is in equilibrium, the resultant force on the rod will be zero,
and the sum of the moments about any point will also be zero.
This PowerPoint produced by R.Collins; ©ZigZag Education 2010