Higher Physics: Topic 2 – Mechanics
Homework 1 MARKING SHEME
SECTION A
Question
1.
Which of the following pairs contains one vector quantity and one scalar
quantity?
A. Force, kinetic energy
2.
Marks
(1)
1 mark
A car of mass 900 kg pulls a caravan of mass 400 kg along a straight horizontal
road with an acceleration of 2 ms-2.
(1)
Assuming that the frictional forces on the caravan are negligible, calculate the
tension in the coupling between the car and the caravan.
m = 400 kg
a = 2 ms-2
F =F
F = ma
F = 400 x 2
F = 800 N
3.
1 mark
The diagram shows two trolleys X and Y about to collide. The momentum of
each trolley before impact is given.
20 kg ms-1
12 kg ms-1
Trolley X
Trolley Y
(1)
After the collision, the trolleys travel in opposite directions.
The magnitude of the momentum of trolley X is 2 kg ms-1.
What is the corresponding magnitude of the momentum of trolley Y in
kg ms-1?
Before Collision
pX = 20 kg ms-1
pY = -12 kg ms-1
After Collision
pX = -2 kg ms-1
pY = pY
momentum before the collision = momentum after the collision
pX + pY = pX + pY
20 + (-12) = (-2) + pY
8 = (-2) + pY
pY = 8 – (-2)
pY = 10 kg ms-1
4.
An inelastic collision takes place between a moving object and a stationary
object. Both objects have the same mass. In this situation, state which of the
following are conserved: momentum, kinetic energy and total energy.
Momentum is conserved, kinetic energy is not conserved and total energy
is conserved.
1 mark
(1)
1 mark
1
Higher Physics: Topic 2 – Mechanics
Question
5.
Homework 1 MARKING SHEME
A rocket of mass 5 kg is travelling horizontally with a speed of 200 ms -1 when
it explodes into two parts. One part of mass 3 kg continues in the original
direction with a speed of 100 ms-1. The other part also continues in this same
direction. Calculate its speed.
Marks
(1)
m = 5 kg
u = 200 ms-1
m1 = 3 kg
m2 = m – m1 = 5 – 3 = 2 kg
v1 = 100 ms-1
v2 = v2
momentum before the explosion = momentum after the explosion
mu = m1v1 + m2v2
(5 x 200) = (3 x 100) + (2 x v2)
1000 = 300 + 2v2
1000 – 300 = 2v2
700 = 2v2
v2 = 700/2
v2 = 350 ms-1
6.
A rocket of mass 200 kg accelerates vertically upwards from the surface of a
planet at 2 ms-2.
1 mark
(1)
The gravitational field strength on the planet is 4 N kg-1.
What is the size of the forces being supplied by the rocket’s engine?
m = 200 kg
a = 2 ms-2
g = 4 N kg-1
F=F
The forces supplied by the rockets engine must be equal to the rockets
weight in order to overcome it plus the force required to produce an
acceleration of 2 ms-2.
W = mg
W = 200 x 4
W = 800 N
F = ma
F = 200 x 2
F = 400 N
Force supplied by the rockets engine = F + W
Force supplied by the rockets engine = 400 + 800
Force supplied by the rockets engine = 1200 N
1 mark
2
Higher Physics: Topic 2 – Mechanics
Homework 1 MARKING SHEME
SECTION B
Question
7.
(a)
Mark
A box of mass 18 kg is at rest on a horizontal frictionless surface.
A force of 4 N is applied to the box at an angle of 26o to the horizontal.
4N
26o
(i)
Show that the horizontal component of this force is 3.6 N.
(2)
½ mark
½ mark
½ mark each
FH = Rcos
FH = 4 x cos 26o
FH = 3.6 N
(ii)
Calculate the acceleration of the box along the horizontal surface.
(2)
FH = 3.6 N
m = 18 kg
a =a
½ mark
½ mark
½ mark each
a = FH/m
a = 3.6/18
a = 0.2 ms-2
(iii)
Calculate the horizontal distance travelled by the box in a time of 7 s.
(2)
t=7s
a = 0.2 ms-2
u = 0 ms-1
s=s
s = ut + ½at2
s = (0 x 7) + (½ x 0.2 x 72)
s = 0 + 4.9
s = 4.9 m
(b)
½ mark
½ mark
½ mark each
(1)
The box is replaced at rest at its starting position.
The force of 4 N is now applied to the box at an angle of less that 26 o to the
horizontal.
4N
angle less that 26o
The force is applied for a time of 7 s as before.
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Higher Physics: Topic 2 – Mechanics
Question
7. (cont)
(b)
Homework 1 MARKING SHEME
Mark
How does the distance travelled by the box compare with your answer to part
(a) (iii)?
You must justify your answer.
FH = Rcos
FH = 4 x cos 10o
FH = 3.9 N
FH = 3.9 N
m = 18 kg
a =a
a = FH/m
a = 3.9/18
a = 0.22 ms-2
t=7s
a = 0.22 ms-2
u = 0 ms-1
s=s
s = ut + ½at2
s = (0 x 7) + (½ x 0.22 x 72)
s = 0 + 5.4
s = 5.4 m
The distance travelled will be greater because the horizontal component of
the force is greater causing the acceleration to be greater.
1 mark
8.
Beads of liquid moving at high speed are used to move threads in modern
weaving machines.
In one design of machine, beads of water are accelerated by jets of air as shown
in the diagram.
jet of air
narrow tube
bead of water
Each bead has a mass of 2.5 x 10-5 kg.
When designing the machine, it was estimated that each bead of water would
start from rest and experience a constant unbalanced force of 0.5 N for a time
of 3 ms.
4
Higher Physics: Topic 2 – Mechanics
Question
8. (cont.)
(a)
(i)
Homework 1 MARKING SHEME
Mark
Calculate:
The impulse on a bead of water;
(2)
F = 0.5 N
t = 3 ms = 3 x 10-3 s
Impulse = Impulse
Impulse = Ft
Impulse = 0.5 x (3 x 10-3)
Impulse = 0.0015 N s = 1.5 x 10-3 N s
(ii)
½ mark
½ mark
½ mark each
The speed of the bead as it emerges from the tube.
(4)
u = 0 ms-1
F = 0.5 N
t = 3 x 10-3 s
m = 2.5 x 10-5 kg
v=v
a = F/m
a = 0.5/2.5 x 10-5
a = 20000 ms-2 = 2 x 104 ms-2
½ mark
½ mark
½ mark each
v = u + at
v = 0 + (20000 x (3 x 10-3))
v = 60 ms-1
½ mark
½ mark
½ mark each
In practice the force on a bead varies.
The following graph shows how the actual unbalanced force exerted on each
bead of water varies with time.
(8)
1
Force/N
(b)
0.5
0
0
3
time/ms
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Higher Physics: Topic 2 – Mechanics
Question
8. (cont.)
(b)
Homework 1 MARKING SHEME
Mark
Use information from this graph to show that the bead leaves the tube with a
speed equal to half of the value calculated in part (ii).
Impulse = area under the F-t graph
Impulse = ½ x 0.5 x (3 x 10-3)
Impulse = 7.5 x 10-4 N s
½ mark
½ mark
½ mark each
F = Impulse/t
F = 7.5 x 10-4/3 x 10-3
F = 0.25 N
½ mark
½ mark
½ mark each
u = 0 ms-1
F = 0.25 N
t = 3 x 10-3 s
m = 2.5 x 10-5 kg
v=v
a = F/m
a = 0.25/2.5 x 10-5
a = 10000 ms-2
½ mark
½ mark
½ mark each
v = u + at
v = 0 + (10000 x (3 x 10-3))
v = 30 ms-1
½ mark
½ mark
½ mark each
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