Explain, with examples, how io identifu a Newton's
third law pair of forces.
A man is standing on a chair resting on the ground.
a) Draw 3 free-body force diagrams, for
i) the man.
ii) the chair, and
iii) the Earth,
showing all the forces acting on each object.
b) Show clearly which of these forces are Newton
pairs of equalbut opposite forces.
"L234time(s)
The graph shows the force acting on a 3.0 kg mass
over a period of 4.0 s.
a) Use the graph to calculate the total impulse.
b) What change in momenhlm does this impulse
produce?
c)
The diagram shows the forces acting on a cyclist
freewheeling down a hill. Her total mass, including
the bicycle is 79 kg. She accelerates from rest to
6.0 m s*l in 3.0 s.
a) What is her acceleration?
b) Calculate the weight, W.
c) By applying Newton's second law along the
slope, find the drag force, D.
d) After a few seconds she reaches a steady speed
down the slope. What is the drag force now?
Why has it changed?
,t"3" During a hailstorm 0.54 kg of hail-stones fall on to
a flat roof in 6.0 s without rebounding. The hail
hits the roof at 12 m s-1. Calculate:
a) the change of momenfum each second,
b) the force exerted on the roof.
'it
ttr
"
il$" During a serve, a tennis racket is in contact with the
ball for 25 ms. The 60 g ball leaves the racket with
a horizontalvelocity of 31 m s-i. If its iniiial
horizontal velocity was zero, calculate:
a) the change in momentum of the ball,
b) the impulse from the racket,
c) ihe average force exerted on the ball.
n i-.
A car moving at 25 m s-1 collides with a wall. The
driver of mass 65 kg is brought to rest by his seat
belt in 0.20 s. Calculate:
a) the driver's change in momenfum,
b) the average force exerted by the seat belt on the
driver.
c) How many times
greater is this force than the
driver's own weight?
A glider of mass 200 g is moving at 0.60 m 11
along an air-track. It collides with a second
stationary glider of mass 250 g.
If the gliders stick together on impact, calculate
their new combined speed.
":,,U"
A car of mass 1500 kg havelling at 12 m s-l
collides head-on with a lorry moving at 20 m s-1.
The lorry has a mass of 9000 kg. If the collision
reduces the speed of the lorry to 15 m s-1, what is
the car's velocity after the impact?
In which direction is this?
1.1i,
A snooker ballmoving at2.0 m s-l collides headon with an identical but stationary bal[. After the
collision ihe second ball moves off at 1.5 m s-1.
What is the new velocity of the first ball?
'9, An empty lift has a mass of 1200k9. It is supported
by a steelcable attached to the roof. Draw a
diagram showing the two forces acting on the lift.
Find the tension in the lift cable when the lift is:
a) ascending at a steady speed,
b) accelerating upwards at 1.5 ffi s-2,
c) accelerating downwards at 2.0 m s-2.
If the mass was initially at rest, what is its
velocity after the 4 s?
3.7. A rocket of total mass 3500 kg is moving at
250 m s-l ihrough space. When the booster
rockets are fired 1200 kg of burnt fuel is ejected
from the back of the rocket at 20 m s-I.
What is ihe new speed of the rocket?
1.8
"
A cannon fires a cannonball of mass 55 kg at
35 m 11. The cannon recoils at2.5 m s-1.
a) Whai is the mass of the cannon?
b) If the cannonball becomes embedded in a target
of mass 600 kg, at what speed does the target
move immediately after impact?
answers
6.0 m s-I
,Au
o
8. a,o=-f = 3-0, :z.ums-'
b) W=mg:790N
c)
Force down slope
=
(W cos60o
UsingF = ma, 79Ocos60o-D
... D=240N (2s.f.)
9.
- D)
=79kg x 2.0ms-2
d) Force up slope = force down slope
.'. D - 790cos60o = 400N (2s.f.)
See Example
7.
4Tension,
r
TI
Lll
a)
b)
r= w=ms=t/#"d*fl't
T-
W = mo .'. T- 12000N = 1200kg x 1.5 m s-z
.'. T:14000N (2s.f.)
c) W -T : ma .'. 12000N-T = 1200kg x 2.0 ms-2
.'. T=9600N
10. a) p = mt) = 0.060kg x 31ms-1 :1.9kgm11 (2s.f.)
b) AP = imPulse : 1.9 N s (2 s.f.)
c) Force = impulse + time = 9.3 N
11. a) Ap : mLu = 65 x 25 = 1625 = 1600 kg m sr 12 s.f.)
:
= + : 81ooN (2
ffiF
c) W = m9: 650 N .'. force 12W
b) Force
s.r.)
-
12. See Example 11.
a) Impulse = area under graph : 10 N s
b) Ap = impulse = 10 kg m r1
c) u = p + m = 10kgms-l + 3.0s : 3.3 m11
13. See Example 13.
a) mass per second = 0.09 kg
.'. Ap per second = mAu = 1.1 kg m s-l (2 s.f.)
b) F
= 1.1N
(2 s.f.)
14. momentum before = momenfum after
x 0.60) + 0 = (0.20 + 0.25) x
.'.u=0.27mr1
(0.20
15. See Example
u
14.
momenhrm before : momentum after
(9000 x 20) - (1500 x 72) = (9000 x 15)
.'. u = 18 m s-1 in direction of lorry
16. Let mass of each ball = m
momentum before - momentum after
+
(1500
(mx2)*0:mu+(mx1.5)
.'.
u = 0.5 m s-l in the direction of the first ball
17. See Example
15.
momentum before = momenfum after
(3500 x 250) = (2300 x u) - (1200 x 20)
.'.u=390m11
18. See Example
L5.
a) momenhrm before = momentum after
0=(55x35)-(mx2.5\
... m-770ks
b) momenfum before = momentum after
(55
x 35) =
(655 kg
x
u)
.'. u=2.9ms-l
x
u)