Download 9702 Dynamics Paper 2 with Answers

9702/2 M/J02
4
1
A steel ball of mass 73 g is held 1.6 m above a horizontal steel plate, as illustrated in Fig. 4.1.
steel ball
mass 73 g
/
m
o
c
.
it
1.6 m
w
s
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h
c
i
s
steel
plate
y
h
p
.
w
Fig. 4.1
The ball is dropped from rest and it bounces on the plate, reaching a height h.
(a) Calculate the speed of the ball as it reaches the plate.
w
w
/: /
tt p
h
speed = .......................................... m s–1
[2]
(b) As the ball loses contact with the plate after bouncing, the kinetic energy of the ball is
90% of that just before bouncing. Calculate
(i)
the height h to which the ball bounces,
h = ............................................... m
2
(ii)
the speed of the ball as it leaves the plate after bouncing.
/
m
o
c
.
it
speed = .......................................... m s–1
(c) Using your answers to (a) and (b), determine the change in momentum of the ball
during the bounce.
c
i
s
y
h
p
.
w
w
w
change = ............................................. N s
[3]
/: /
(d) With reference to the law of conservation of momentum, comment on your answer
to (c).
tt p
..........................................................................................................................................
..........................................................................................................................................
h
...............................................................................................................................
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w
s
[4]
9702/2/O/N/02
3
3
A ball falls from rest onto a flat horizontal surface. Fig. 3.1 shows the variation with time t of
the velocity v of the ball as it approaches and rebounds from the surface.
/
m
o
5
v / m s –1
4
w
s
2
1
0
0.1
0
0.2
0.3
y
h
c
i
s
-1
0.4
0.5
0.6
0.7
t/s
p
.
w
-2
-3
w
w
-4
/: /
tt p
Fig. 3.1
Use data from Fig. 3.1 to determine
(a) the distance travelled by the ball during the first 0.40 s,
h
distance = ……………………………………. m [2]
(b) the change in momentum of the ball, of mass 45 g, during contact of the ball with the
surface,
change = ………………………………….. N s [4]
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c
.
it
3
4
(c) the average force acting on the ball during contact with the surface.
/
m
o
c
.
it
w
s
force = ……………………………………. N [2]
c
i
s
y
h
p
.
w
/: /
w
w
h
tt p
5
9702/02/M/J/04
4
A ball has mass m. It is dropped onto a horizontal plate as shown in Fig. 4.1.
/
m
o
plate
v
w
s
c
i
s
Fig. 4.1
y
h
Just as the ball makes contact with the plate, it has velocity v, momentum p and kinetic
energy Ek.
(a) (i)
p
.
w
Write down an expression for momentum p in terms of m and v.
...................................................................................................................................
(ii)
Hence show that the kinetic energy is given by the expression
/: /
w
w
tt p
Ek =
p2
.
2m
h
[3]
(b) Just before impact with the plate, the ball of mass 35 g has speed 4.5 m s–1. It bounces
from the plate so that its speed immediately after losing contact with the plate is
3.5 m s–1. The ball is in contact with the plate for 0.14 s.
Calculate, for the time that the ball is in contact with the plate,
(i)
the average force, in addition to the weight of the ball, that the plate exerts on the
ball,
magnitude of force = .................................... N
direction of force = ........................................
[4]
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c
.
it
6
(ii)
the loss in kinetic energy of the ball.
/
m
o
w
s
(c) State and explain whether linear momentum is conserved during the bounce.
c
i
s
..........................................................................................................................................
..........................................................................................................................................
y
h
..........................................................................................................................................
p
.
w
.................................................................................................................................... [3]
/: /
w
w
h
tt p
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c
.
it
loss = ....................................... J [2]
7
9702/02/O/N/04
3
A girl stands at the top of a cliff and throws a ball vertically upwards with a speed of 12 m s–1,
as illustrated in Fig. 3.1.
path of
ball
/
m
o
w
s
c
i
s
y
h
Fig. 3.1
At the time that the girl throws the ball, her hand is a height h above the horizontal ground at
the base of the cliff.
The variation with time t of the speed v of the ball is shown in Fig. 3.2.
p
.
w
20
w
w
–1
v/ms
/: /
10
h
tt p
0
0
1.0
2.0
–10
–20
–30
–40
Fig. 3.2
3.0
4.0
5.0
t/s
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c
.
it
h
8
Speeds in the upward direction are shown as being positive. Speeds in the downward
direction are negative.
(a) State the feature of Fig. 3.2 that shows that the acceleration is constant.
/
.................................................................................................................................... [1]
m
o
(b) Use Fig. 3.2 to determine the time at which the ball
(i)
reaches maximum height,
c
.
it
time = ………………………………. s
hits the ground at the base of the cliff.
w
s
time = ………………………………. s
[2]
c
i
s
(c) Determine the maximum height above the base of the cliff to which the ball rises.
y
h
p
.
w
/: /
w
w
tt p
height = …………………………… m [3]
(d) The ball has mass 250 g. Calculate the magnitude of the change in momentum of the
ball between the time that it leaves the girl’s hand to time t = 4.0 s.
h
change = …………………………… N s [3]
(e) (i)
State the principle of conservation of momentum.
...................................................................................................................................
...................................................................................................................................
............................................................................................................................. [2]
(ii)
Comment on your answer to (d) by reference to this principle.
...................................................................................................................................
...................................................................................................................................
...................................................................................................................................
............................................................................................................................. [3]
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(ii)
9702/02/M/J/05
3
9
A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block is
suspended from strings so that it is free to move in a vertical plane.
The bullet buries itself in the block. The block and bullet rise together through a vertical
distance of 8.6 cm, as shown in Fig. 3.1.
/
m
o
wood
block
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c
.
it
w
s
c
i
s
bullet
y
h
8.6 cm
p
.
w
Fig. 3.1
(a) (i)
w
w
Calculate the change in gravitational potential energy of the block and bullet.
/: /
tt p
h
change = ............................................. J [2]
(ii)
Show that the initial speed of the block and the bullet, after they began to move off
together, was 1.3 m s–1.
[1]
10
(b) Using the information in (a)(ii) and the principle of conservation of momentum,
determine the speed of the bullet before the impact with the block.
/
m
o
c
.
it
speed = ....................................... m s–1 [2]
w
s
Calculate the kinetic energy of the bullet just before impact.
c
i
s
y
h
p
.
w
kinetic energy = .............................................. J [2]
(ii)
w
w
State and explain what can be deduced from your answers to (c)(i) and (a)(i) about
the type of collision between the bullet and the block.
/: /
...................................................................................................................................
tt p
...................................................................................................................................
h
...............................................................................................................................[2]
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(c) (i)
9702/02/M/J/06
1
11
(a) Derive the SI base unit of force.
(b) A spherical ball of radius r experiences a resistive force F due to the air as it moves
through the air at speed v. The resistive force F is given by the expression
F = crv,
where c is a constant.
Derive the SI base unit of the constant c.
SI base unit of c = ………………………………… [1]
(c) The ball is dropped from rest through a height of 4.5 m.
(i)
Assuming air resistance to be negligible, calculate the final speed of the ball.
speed = …………………………… m s–1 [2]
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SI base unit of force = ………………………………… [1]
12
(ii)
The ball has mass 15 g and radius 1.2 cm.
The numerical value of the constant c in the equation in (b) is equal to 3.2 × 10–4
when measured using the SI system of units.
Show quantitatively whether the assumption made in (i) is justified.
m
o
c
.
it
9702/02/O/N/05
3
/
[3]
A stone on a string is made to travel along a horizontal circular path, as shown in Fig. 3.1.
c
i
s
y
h
path of
stone
w
w
/: /
tt p
p
.
w
stone
Fig. 3.1
The stone has a constant speed.
h
(a) Define acceleration.
..........................................................................................................................................
..................................................................................................................................... [1]
(b) Use your definition to explain whether the stone is accelerating.
..........................................................................................................................................
..........................................................................................................................................
..................................................................................................................................... [2]
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w
s
13
(c) The stone has a weight of 5.0 N. When the string makes an angle of 35° to the vertical,
the tension in the string is 6.1 N, as illustrated in Fig. 3.2.
/
6.1 N
m
o
35°
w
s
c
i
s
y
h
5.0 N
p
.
w
Fig. 3.2
Determine the resultant force acting on the stone in the position shown.
w
w
/: /
tt p
h
magnitude of force = ……………..……………………. N
direction of force….………………..………………….. [4]
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c
.
it
9702/02/O/N/06
3
14
Francium-208 is radioactive and emits α-particles with a kinetic energy of 1.07 × 10–12 J to
form nuclei of astatine, as illustrated in Fig. 3.1.
/
astatine
nucleus
francium nucleus
before decay
m
o
w
s
Fig. 3.1
c
i
s
(a) State the nature of an α-particle.
y
h
..........................................................................................................................................
..................................................................................................................................... [1]
p
.
w
(b) Show that the initial speed of an α-particle after the decay of a francium nucleus is
approximately 1.8 × 107 m s–1.
w
w
/: /
tt p
h
[2]
(c) (i)
State the principle of conservation of linear momentum.
...................................................................................................................................
...................................................................................................................................
.............................................................................................................................. [2]
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c
.
it
- particle
(ii)
15
The Francium-208 nucleus is stationary before the decay. Estimate the speed of
the astatine nucleus immediately after the decay.
/
m
o
w
s
c
i
s
speed = ………………………… m s–1 [3]
(d) Close examination of the decay of the francium nucleus indicates that the astatine
nucleus and the α-particle are not ejected exactly in opposite directions.
y
h
Suggest an explanation for this observation.
p
.
w
..........................................................................................................................................
..........................................................................................................................................
w
w
..................................................................................................................................... [2]
/: /
h
tt p
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c
.
it
9702/21/M/J/09
2
16
A ball B of mass 1.2 kg travelling at constant velocity collides head-on with a stationary ball S
of mass 3.6 kg, as shown in Fig. 2.1.
/
v
ball B
ball S
mass 1.2 kg
mass 3.6 kg
m
o
w
s
Fig. 2.1
c
i
s
Frictional forces are negligible.
The variation with time t of the velocity v of ball B before, during and after colliding with ball S
is shown in Fig. 2.2.
y
h
p
.
w
+4
+3
v / m s–1
w
w
+2
/: /
+1
0
tt p
0
–1
h
0.1
0.2
0.3
0.4
0.5
t/s
–2
Fig. 2.2
(a) State the significance of positive and negative values for v in Fig. 2.2.
..........................................................................................................................................
.................................................................................................................................... [1]
(b) Use Fig. 2.2 to determine, for ball B during the collision with ball S,
(i)
the change in momentum of ball B,
change in momentum = .......................................... N s [3]
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.
it
17
(ii)
the magnitude of the force acting on ball B.
/
m
o
c
.
it
w
s
force = ............................................. N [3]
y
h
p
.
w
/: /
w
w
speed = ....................................... m s–1 [2]
(d) Using your answer in (c) and information from Fig. 2.2, deduce quantitatively whether
the collision is elastic or inelastic.
tt p
h
..........................................................................................................................................
.................................................................................................................................... [2]
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c
i
s
(c) Calculate the speed of ball S after the collision.
18
9702/22/M/J/09
2
An experiment is conducted on the surface of the planet Mars.
A sphere of mass 0.78 kg is projected almost vertically upwards from the surface of the
planet. The variation with time t of the vertical velocity v in the upward direction is shown in
Fig. 2.1.
10
m
o
v /m s-1
0
1
w
s
c
i
s
2
3
4 t /s
y
h
–5
p
.
w
–10
w
w
Fig. 2.1
The sphere lands on a small hill at time t = 4.0 s.
/: /
(a) State the time t at which the sphere reaches its maximum height above the planet’s
surface.
t = .............................................. s [1]
tt p
h
(b) Determine the vertical height above the point of projection at which the sphere finally
comes to rest on the hill.
height = ............................................. m [3]
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c
.
it
5
0
/
19
(c) Calculate, for the first 3.5 s of the motion of the sphere,
(i)
the change in momentum of the sphere,
/
m
o
w
s
c
i
s
change in momentum = ...........................................N s [2]
(ii)
the force acting on the sphere.
y
h
p
.
w
w
w
/: /
tt p
force = ..............................................N [2]
(d) Using your answer in (c)(ii),
h
(i)
(ii)
state the weight of the sphere,
weight = ..............................................N [1]
determine the acceleration of free fall on the surface of Mars.
acceleration = ........................................ m s–2 [2]
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c
.
it
9702/21/O/N/09
3
20
A stationary nucleus of mass 220u undergoes radioactive decay to produce a nucleus D of
mass 216u and an α-particle of mass 4u, as illustrated in Fig. 3.1.
/
nucleus
before decay
m
o
220u
α-particle
c
i
s
216u
y
h
4u
initial kinetic energy
1.0 × 10–12 J
Fig. 3.1
p
.
w
The initial kinetic energy of the α-particle is 1.0 × 10–12 J.
(a) (i)
State the law of conservation of linear momentum.
w
w
..................................................................................................................................
/: /
..................................................................................................................................
tt p
............................................................................................................................ [2]
h
(ii)
Explain why the initial velocities of the nucleus D and the α-particle must be in
opposite directions.
..................................................................................................................................
..................................................................................................................................
............................................................................................................................ [2]
(b) (i)
Show that the initial speed of the α-particle is 1.7 × 107 m s–1.
[2]
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w
s
nucleus D
after decay
c
.
it
21
(ii)
Calculate the initial speed of nucleus D.
/
m
o
c
.
it
w
s
speed = ...................................... m s–1 [2]
y
h
p
.
w
/: /
w
w
h
tt p
deceleration = ...................................... m s–2 [2]
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c
i
s
(c) The range in air of the emitted α-particle is 4.5 cm.
Calculate the average deceleration of the α-particle as it is stopped by the air.
9702/22/M/J/10
3
(a) (i)
22
Define force.
..................................................................................................................................
/
.............................................................................................................................. [1]
(ii)
m
o
State Newton’s third law of motion.
c
.
it
..................................................................................................................................
..................................................................................................................................
w
s
..................................................................................................................................
(b) Two spheres approach one another along a line joining their centres, as illustrated in
Fig. 3.1.
y
h
p
.
w
sphere
A
w
w
/: /
sphere
B
Fig. 3.1
When they collide, the average force acting on sphere A is FA and the average force
acting on sphere B is FB.
tt p
h
The forces act for time tA on sphere A and time tB on sphere B.
(i)
State the relationship between
1. FA and FB,
.............................................................................................................................. [1]
2. tA and tB.
.............................................................................................................................. [1]
(ii)
Use your answers in (i) to show that the change in momentum of sphere A is equal
in magnitude and opposite in direction to the change in momentum of sphere B.
..................................................................................................................................
.............................................................................................................................. [1]
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c
i
s
.............................................................................................................................. [3]
23
(c) For the spheres in (b), the variation with time of the momentum of sphere A before,
during and after the collision with sphere B is shown in Fig. 3.2.
/
15
momentum
to right / N s
m
o
10
w
s
5
c
i
s
0
y
h
sphere B
p
.
w
–-5
5
w
w
–10
-10
/: /
–15
-15
h
tt p
time
Fig. 3.2
The momentum of sphere B before the collision is also shown on Fig. 3.2.
Complete Fig. 3.2 to show the variation with time of the momentum of sphere B during
and after the collision with sphere A.
[3]
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c
.
it
sphere A
9702/21/O/N/10
3
24
(a) State the relation between force and momentum.
.................................................................................................................................... [1]
(b) A rigid bar of mass 450 g is held horizontally by two supports A and B, as shown in
Fig. 3.1.
m
o
ball
45 cm
c
i
s
25 cm
B
y
h
Fig. 3.1
The support A is 45 cm from the centre of gravity C of the bar and support B is 25 cm
from C.
p
.
w
A ball of mass 140 g falls vertically onto the bar such that it hits the bar at a distance of
50 cm from C, as shown in Fig. 3.1.
The variation with time t of the velocity v of the ball before, during and after hitting the
bar is shown in Fig. 3.2.
w
w
/: /
tt p
velocity
downwards
/ m s–1
h
4
2
0
0
0.2
0.4
0.6
0.8
1.0
1.2
time / s
–2
–4
–6
Fig. 3.2
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50 cm
c
.
it
w
s
A
C
6
/
25
For the time that the ball is in contact with the bar, use Fig. 3.2
(i)
to determine the change in momentum of the ball,
/
m
o
w
s
change = .................................. kg m s–1 [2]
(ii)
c
i
s
to show that the force exerted by the ball on the bar is 33 N.
y
h
p
.
w
w
w
/: /
[1]
(c) For the time that the ball is in contact with the bar, use data from Fig. 3.1 and (b)(ii) to
calculate the force exerted on the bar by
tt p
(i)
the support A,
h
force = ............................................ N [3]
(ii)
the support B.
force = ............................................ N [2]
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.
it
9702/22/M/J/11
2
26
(a) A sphere of radius R is moving through a fluid with constant speed v. There is a frictional
force F acting on the sphere, which is given by the expression
F = 6πDRv
/
where D depends on the fluid.
(i)
m
o
Show that the SI base units of the quantity D are kg m–1 s–1.
w
s
c
i
s
(ii)
y
h
p
.
w
A raindrop of radius 1.5 mm falls vertically in air at a velocity of 3.7 m s–1. The value
of D for air is 6.6 × 10–4 kg m–1 s–1. The density of water is 1000 kg m–3.
Calculate
1.
w
w
the magnitude of the frictional force F,
/: /
tt p
h
[3]
2.
F = ............................................. N [1]
the acceleration of the raindrop.
acceleration = ........................................ m s–2 [3]
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.
it
27
(b) The variation with time t of the speed v of the raindrop in (a) is shown in Fig. 2.1.
/
m
o
v
w
s
0
0
y
h
Fig. 2.1
(i)
c
i
s
t
p
.
w
State the variation with time of the acceleration of the raindrop.
..................................................................................................................................
w
w
..................................................................................................................................
/: /
..................................................................................................................................
..................................................................................................................................
tt p
............................................................................................................................. [3]
h
(ii)
A second raindrop has a radius that is smaller than that given in (a). On Fig. 2.1,
sketch the variation of speed with time for this second raindrop.
[2]
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c
.
it
9702/23/M/J/11
3
28
A helicopter has a cable hanging from it towards the sea below, as shown in Fig. 3.1.
/
helicopter
m
o
cable
w
s
sea
Fig. 3.1
c
i
s
y
h
A man of mass 80 kg rescues a child of mass 50.5 kg. The two are attached to the cable
and are lifted from the sea to the helicopter. The lifting process consists of an initial uniform
acceleration followed by a period of constant velocity and then completed by a final uniform
deceleration.
p
.
w
(a) Calculate the combined weight of the man and child.
w
w
/: /
tt p
weight = ............................................. N [1]
(b) Calculate the tension in the cable during
h
(i)
the initial acceleration of 0.570 m s–2,
tension = ............................................. N [2]
(ii)
the period of constant velocity of 2.00 m s–1.
tension = ............................................. N [1]
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c
.
it
29
(c) During the final deceleration the tension in the cable is 1240 N. Calculate this
deceleration.
/
m
o
deceleration = ........................................ m s–2 [2]
(d) (i)
1.
moving with uniform acceleration,
w
s
c
i
s
time = .............................................. s [1]
y
h
2.
p
.
w
(ii)
time = .............................................. s [1]
The time over which the man and child are moving with constant velocity is 20 s.
On Fig. 3.2, sketch a graph to show the variation with time of the velocity of the
man and child for the complete lifting process.
w
w
/: /
2.0
velocity
/ m s–1
moving with uniform deceleration.
tt p
h
1.0
0
0
5
10
15
20
25
30
35
time / s
Fig. 3.2
[2]
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c
.
it
Calculate the time over which the man and child are
9702/22/M/J/12
3
30
(a) State Newton’s first law.
..........................................................................................................................................
/
...................................................................................................................................... [1]
m
o
(b) A log of mass 450 kg is pulled up a slope by a wire attached to a motor, as shown in
Fig. 3.1.
12°
w
s
c
i
s
Fig. 3.1
y
h
The angle that the slope makes with the horizontal is 12°. The frictional force acting on
the log is 650 N. The log travels with constant velocity.
(i)
p
.
w
With reference to the motion of the log, discuss whether the log is in equilibrium.
..................................................................................................................................
w
w
..................................................................................................................................
..................................................................................................................................
/: /
.............................................................................................................................. [2]
tt p
(ii)
Calculate the tension in the wire.
h
tension = ............................................. N [3]
(iii)
State and explain whether the gain in the potential energy per unit time of the log is
equal to the output power of the motor.
..................................................................................................................................
..................................................................................................................................
..................................................................................................................................
.............................................................................................................................. [2]
[email protected] Cell No. 03467030039
c
.
it
motor
wire
log
9702/23/M/J/12
2
31
A motor drags a log of mass 452 kg up a slope by means of a cable, as shown in Fig. 2.1.
10.0
m
cable
start and finish
position
of log
m
o
P
w
s
Fig. 2.1
The slope is inclined at 14.0° to the horizontal.
c
i
s
(a) Show that the component of the weight of the log acting down the slope is 1070 N.
y
h
p
.
w
[1]
w
w
(b) The log starts from rest. A constant frictional force of 525 N acts on the log. The log
accelerates up the slope at 0.130 m s–2.
(i)
/: /
Calculate the tension in the cable.
tt p
h
tension = ............................................. N [3]
(ii)
The log is initially at rest at point S. It is pulled through a distance of 10.0 m to
point P.
Calculate, for the log,
1.
the time taken to move from S to P,
time = .............................................. s [2]
2.
the magnitude of the velocity at P.
velocity = ........................................ m s
[email protected] Cell No. 03467030039
c
.
it
14.0°
S
/
motor
32
(c) The cable breaks when the log reaches point P. On Fig. 2.2, sketch the variation with
time t of the velocity v of the log. The graph should show v from the start at S until the
log returns to S.
[4]
/
m
o
v
w
s
0
0
c
i
s
t
y
h
p
.
w
2
Fig. 2.2
w
w
9702/22/O/N/12
(a) State Newton’s second law.
/: /
..........................................................................................................................................
...................................................................................................................................... [1]
tt p
(b) A ball of mass 65 g hits a wall with a velocity of 5.2 m s–1 perpendicular to the wall. The
ball rebounds perpendicularly from the wall with a speed of 3.7 m s–1. The contact time
of the ball with the wall is 7.5 ms.
h
Calculate, for the ball hitting the wall,
(i)
the change in momentum,
change in momentum = ........................................... N s [2]
(ii)
the magnitude of the average force.
force = ............................................. N [1]
[email protected] Cell No. 03467030039
c
.
it
33
(c) (i)
For the collision in (b) between the ball and the wall, state how the following apply:
1. Newton’s third law,
..................................................................................................................................
/
m
o
..................................................................................................................................
c
.
it
..................................................................................................................................
.............................................................................................................................. [2]
c
i
s
..................................................................................................................................
.............................................................................................................................. [1]
(ii)
y
h
State, with a reason, whether the collision is elastic or inelastic.
p
.
w
..................................................................................................................................
.............................................................................................................................. [1]
9702/23/M/J/13
3
(a) (i)
w
w
State the principle of conservation of momentum.
/: /
..................................................................................................................................
..................................................................................................................................
tt p
.............................................................................................................................. [2]
h
(ii)
State the difference between an elastic and an inelastic collision.
.............................................................................................................................. [1]
(b) An object A of mass 4.2 kg and horizontal velocity 3.6 m s–1 moves towards object B as
shown in Fig. 3.1.
A
B
3.6 m s–1
1.2 m s–1
1.5 kg
4.2 kg
before collision
Fig. 3.1
Object B of mass 1.5 kg is moving with a horizontal velocity of 1.2 m s–1 towards
object A.
[email protected] Cell No. 03467030039
w
s
2. the law of conservation of momentum.
34
The objects collide and then both move to the right, as shown in Fig. 3.2.
A
B
4.2 kg
v
1.5 kg
Fig. 3.2
Calculate the velocity v of object A after the collision.
m
o
w
s
c
i
s
y
h
velocity = ........................................ m s–1 [3]
(ii)
p
.
w
Determine whether the collision is elastic or inelastic.
/: /
w
w
h
tt p
[3]
[email protected] Cell No. 03467030039
c
.
it
Object A has velocity v and object B has velocity 3.0 m s–1.
(i)
/
3.0 m s–1
after collision
35
9702/22/M/J/14
3
(a) State Newton’s first law of motion.
...................................................................................................................................................
/
.............................................................................................................................................. [1]
m
o
(b) A box slides down a slope, as shown in Fig. 3.1.
box
20°
w
s
c
i
s
horizontal
y
h
Fig. 3.1
The angle of the slope to the horizontal is 20°. The box has a mass of 65 kg. The total resistive
force R acting on the box is constant as it slides down the slope.
(i)
p
.
w
State the names and directions of the other two forces acting on the box.
1. ........................................................................................................................................
w
w
2. ........................................................................................................................................
[2]
(ii)
/: /
The variation with time t of the velocity v of the box as it moves down the slope is shown
in Fig. 3.2.
tt p
8.0
h
6.0
v / m s–1
4.0
2.0
0
0
1.0
Fig. 3.2
t /s
2.0
[email protected] Cell No. 03467030039
c
.
it
v
36
1. Use data from Fig. 3.2 to show that the acceleration of the box is 2.6
m s–2.
/
.t c
m
o
s
ic
s
y
h
p
resultant force = ...................................................... N [1]
.
w
3. Determine the resistive force R on the box.
w
w
/: /
h
tt p
R = ...................................................... N [3]
[email protected] Cell No. 03467030039
i
w
2. Calculate the resultant force on the box.
[2]
37
/
m
o
c
.
it
[email protected] Cell No. 03467030039
w
s
c
i
s
y
h
p
.
w
w
w
/: /
tt p
h
2
A/AS LEVEL EXAMINATIONS - NOVEMBER 2003
9702
02
hence 6 photographs (‘bald’ answer scores 2 marks only) ........... A1
[3]
(a)
mass: measure of body’s resistance/inertia to changes in
velocity/motion ............................................................................. B1
weight: effect of gravitational field on mass or force of gravity ..... B1
any further comment e.g. mass constant, weight varies/
weight = mg/scalar and vector ..................................................... B1
[3]
(b)
e.g. where gravitational field strength changes
(change) in fluid surrounding body…. 1 each, max 2 ................... B2
[2]
A/AS LEVEL EXAMINATIONS - JUNE 2004
4
(a)
(b)
(i)
(ii)
(i)
(ii)
(c)
9702
(p =) mv
1
2
Ek =
mv
2
algebra leading to
Ek = p2/2m
B1
B1
M1
A0
∆p
= 0.035 (4.5 + 3.5)
OR
a = (4.5 + 3.5)/0.14
-2
= 0.28 N s
= 57.1 m s
OR
F = ma (= 0.035 x 575.1) (allow e.c.f.)
force = ∆ p/ ∆ t (= 0.28/0.14)
= 2.0 N
Note: candidate may add mg = 0.34 N to this answer, deduct 1 mark
upwards
1
2
2
loss =
x 0.035 (4.5 – 3.5 )
2
= 0.14 J
(No credit for 0.282/(2 x 0.035) = 1.12 J)
e.g.
02
plate (and Earth) gain momentum
i.e. discusses a 'system'
equal and opposite to the change for the ball
i.e. discusses force/momentum
so momentum is conserved
i.e. discusses consequence
[3]
C1
C1
A1
B1
[4]
C1
A1
[2]
B1
M1
A1
Total
[3]
[12]
A and AS LEVEL – NOVEMBER 2004
3
(e) (i)
9702
38
2
total/sum momentum before = total/sum momentum after
B1
in any closed system
B1
(ii) either
the system is the ball and Earth
B1
momentum of Earth changes by same amount
B1
/
m
o
but in the opposite direction
or
[2]
B1
Ball is not an isolated system/there is a force on the ball (B1)
.t c
Gravitational force acts on the ball
(B1)
causes change in momentum/law does not apply here
i
w
(B1)
[3]
(if explains in terms of air resistance, allow first mark only)
A and AS LEVEL – June 2005
(a)
s
ic
When a wave (front) is incident on an edge
or an obstacle/slit/gap
Wave ‘bends’ into the geometrical
shadow/changes direction/spreads
d = 1/(750 × 10 )
= 1.33 × 10-6 m
-6
M1
s
y
3
(b) (i)
2
(ii) 1.33 × 10 × sin90° = n × 590 × 10
n = 2 (must be an integer)
h
p
-9
(iii) formula assumes no path difference of light before
entering grating or
there is a path difference before the grating
.
w
(c)
(a)
w
w
/: /
kg m s–2
–1
(b) kg m
tt p
(c)
h
(i)
(ii)
[2]
C1
A1
[2]
C1
A1
[2]
B1
[1]
e.g. lines further apart in second order
lines fainter in second order
(allow any sensible difference: 1 each, max 2)
B2
(if differences stated but without reference to the orders, max 1 mark)
GCE A Level – May/June 2006
1
A1
2
–1
s
v = 2gs
= 2 × 9.8 × 4.5
v = 9.4 m s–1
either
F (= 3.2 × 10–4 × 1.2 × 10–2 × 9.4) = 3.6 × 10–5 N
weight of sphere (= mg = 15 × 10–3 × 9.8) = 0.15 N
3.6 × 10–5 << 0.15, so justified
or
mg = crvT
(M1)
(M1)
terminal speed = 3.8 × 104 m s–1
4
9.4 << 3.8 × 10 , so justified
(A1)
9702
[2]
02
B1
[1]
B1
[1]
C1
A1
[2]
M1
M1
A1
[3]
[email protected] Cell No. 03467030039
5
9702
GCE A/AS LEVEL – November 2005
3
9702
(a)
change in velocity/time (taken)
B1
[1]
(b)
velocity is a vector/velocity has magnitude & direction
direction changing so must be accelerating
B1
B1
[2]
either 6.1 × cos35 = 4.99 N
so no resultant vertical force
6.1 sin35 = 3.5 N
horizontally
B1
B1
B1
B1
[4]
(c)
or scale shown
triangle of correct shape
resultant = 3.5 ± 0.2 N
horizontal ± 3°
9702
helium nucleus OR contains two protons and two neutrons
B1
[1]
(b)
kinetic energy = ½mv2
½ × 4 × 1.66 × 10-27 × v2 = 1.07 × 10-12
v = 1.8 × 107 m s-1
C1
A1
A0
[2]
c
i
s
y
h
(i)
sum of momenta (in any direction) is constant
/ total momemtum is constant
in a closed system / no external force
momentum of francium (= 0) = momentum of α + momentum of astatine
204 × V = 4 × 1.8 × 107
V = 3.5 × 105 m s-1
(nuclei incorrectly identified, 0/3
nuclei correctly identified but incorrect masses, -1 each error)
(ii)
p
.
w
(d)
another particle / photon is emitted
at an angle to the direction of the α-particle
(allow 1 mark for ‘Francium nucleus is not stationary’)
w
w
GCE A/AS LEVEL – May/June 2009
(a)
ball moving in opposite direction (after collision)
(b)
(i)
(d)
[3]
M1
A1
[2]
[2]
21
................................
B1
change in momentum = 1.2 (4.0 + 0.8)
.......................................
(correct values, 1 mark; correct sign {values added}, 1 mark )
= 5.76 N s …(allow 5.8)
.........................
C2
/: /
[1]
[3]
= ∆p / ∆t
or m∆v / ∆t
...................................................
= 5.76 / 0.08 or 1.2 × 4.8 / 0.08
........................................
= 72 N
...............................................................................
C1
C1
A1
[3]
5.76 = 3.6 × V
.....................................................................................
V = 1.6 m s-1
.......................................................................................
C1
A1
[2]
speed of approach = 4.0 m s-1 and
speed of separation = 2.4 m s-1
..............................................
not equal and so inelastic
.......................................................
M1
A1
kinetic energy before = 9.6 J and
kinetic energy after collision = 4.99 J
......................................
kinetic energy after is less / not conserved so inelastic
..........
M1
A1
tt p
force
h
(c)
9702
M1
A1
C1
C1
A1
A1
(ii)
either
or
GCE A/AS LEVEL
2
2
(a)
(c)
2
(B1)
(B1)
(B1)
(B1)
May/June 2009
9702
[2]
22
(a)
2.4 s
(b)
in (b) and (c), allow answers as (+) or (–)
recognises distance travelled as area under graph line …………………………..
height = (½ × 2.4 × 9.0) – (½ × 1.6 × 6.0) ………………………………………..
= 6.0 m (allow 6 m) ……………………………………………………………
(answer 15.6 scores 2 marks
answer 10.8 or 4.8 scores 1 mark)
…………………………………………………………………………………….
A1
[1]
C1
C1
A1
[3]
C1
A1
[2]
= ut – ½at2
= (9 × 4) – ½ × (9 / 2.4) × 42
= 6.0 m
(answer 66 scores 2 marks
answer 36 or 30 scores 1 mark)
alternative solution: s
(c)
(i)
(ii)
(d)
(i)
(ii)
change in momentum
= 0.78 (9.0 + 4.2)
(allow 4.2 ± 0.2) ……………….
= 10.3 N s
(allow 10 N s) ………………………….
force = ∆p / ∆t
or
m∆v / ∆t ……………………………………….
= 10.3 / 3.5 / 0.08
= 2.9 N ………………………………………………………………………
A1
[2]
2.9 N ………………………………………………………………………………..
A1
[1]
g
C1
= weight / mass ……………………………………………………………….
= 2.9 / 0.78
= 3.7 m s–2 ……………………………………………………………………..
C1
A1
[2]
[email protected] Cell No. 03467030039
w
s
GCE A/AS LEVEL - OCT/NOV 2006
/
m
o
c
.
it
allow answer based on centripetal force:
resultant is centripetal force (which is horizontal)
resultant is horizontal component of tension
6.1 sin35 = 3.5 N
horizontally
3
39
2
40
GCE A/AS LEVEL
3
(a)
(i)
(ii)
(b)
[2]
zero momentum before / after decay ..................................................................M1
so α-particle and nucleus D must have momenta in opposite directions
........... A1
[2]
................................................................................ C1
1.7 × 10 × 4u = 216u × V
............................................................................................ A1
V = 3.1 × 105 m s-1
5
-1
(accept 3.2 × 10 m s , do not accept 220 rather than 216)
(i)
(i)
(ii)
(c)
s
ic
9702
B1
[1]
force on body A is equal in magnitude to force on body B (from A) …………M1
forces are in opposite directions ……………………………………………… A1
forces are of the same kind ………………………………………………………A1
[3]
s
y
h
p
1 FA = – FB …………………………………………………………………….
2 t A = t B ………………………………………………………………………
∆p = FA t
A
= – FB t
B
.
w
…………………………………………………………..
graph: momentum change occurs at same times for both spheres ………….
final momentum of sphere B is to the right ……………………………………..
and of magnitude 5 N s ……………………………………………………………
w
w
(a)
force = rate of change of momentum
(b)
(i)
(ii)
(a)
(i)
(ii)
(i)
(ii)
[1]
B1
M1
A1
[3]
C1
A1
[2]
force = 1.33 / 0.04
= 33.3 N
M1
A0
[1]
taking moments about B
(33 × 75) + (0.45 × g × 25) = FA × 20
FA = 129 N
C1
C1
A1
[3]
FB = 33 + 129 + 0.45g
= 166 N
C1
A1
[2]
/: /
= 140 × 10–3 × (5.5 + 4.0)
= 1.33 kg m s–1
9702
22
base units of D:
force: kg m s–2
radius: m
velocity: m s–1
B1
B1
base units of D: [F / (R × v)] kg m s–2 / (m × m s–1)
= kg m–1 s–1
M1
A0
[3]
A1
[1]
C1
M1
A1
[3]
a = g at time t = 0
a decreases (as time increases)
a goes to zero
B1
B1
B1
[3]
Correct shape below original line
sketch goes to terminal velocity earlier
M1
A1
[2]
1.
2.
(b)
B1
21
(allow symbols if defined)
GCE AS/A LEVEL – May/June 2011
2
[1]
[1]
[1]
tt p
h
(i)
9702
B1
B1
B1
∆ρ
(ii)
(c)
[2]
22
force is rate of change of momentum …………………………………………
GCE AS/A LEVEL – October/November 2010
3
i
w
[2]
.............................................................. C1
(1.7 × 107)2 = 2 × deceleration × 4.5 × 10-2
deceleration / a = 3.2 × 1015 m s-2
........................................................................... A1
[2]
(accept calculation based on calculating F = 2.22 × 10-11 N
[Total: 10]
and then use of F = ma)
(ii)
(b)
m
o
.t c
7
/
F = 6π × D × R × v = [6π × 6.6 × 10
= 6.9 × 10–5 N
–4
× 1.5 × 10
–3
mg – F = ma
hence a = g – [F / m]
m = ρ × V = ρ × 4/3 π R3 = (1.4 × 10–5)
a = 9.81 – [6.9 × 10–5] / ρ × 4/3 π × (1.5 × 10–3)3
a = 4.9(3) m s–2
× 3.7]
(9.81 – 4.88)
[email protected] Cell No. 03467030039
(a)
60
either sum / total momentum (of system of bodies) is constant
or
total momentum before = total momentum after
......................................M1
for an isolated system / no (external) force acts on system
............................... A1
GCE AS/A LEVEL – May/June 2010
3
9702
kinetic energy = ½ mv2 .. ................................................................................... C1
1.0 × 10-12 = ½ × 4 × 1.66 × 10-27 × v2
..............................................................M1
v = 1.7 × 107 m s-1 ............................................................................................. A0
(i)
(ii)
(c)
October/November 2009
GCE AS/A LEVEL – May/June 2011
3
(a)
(b)
F = ma
T – 1280 = 130.5 × 0.57
T = 1280 + 74.4 = 1350 N
(i)
(d)
1240 – 1280 = 130.5 × a
a = (–) 0.31 m s–2
(ii)
1.
3.5 s
2.
6.5 s
.t c
basic shape
correct points
(b)
(i)
(ii)
h
p
0, 0.8 s (one of these)
(iii)
0.2, 0.6, 1.0 s (one of these)
.
w
GCE AS/A LEVEL – May/June 2012
2
9702
9702
[2]
/
A1
[1]
C1
A1
[2]
A1
[1]
A1
[1]
M1
A1
[2]
B1
B1
[2]
A1
[1]
A1
[1]
A1
[1]
23
(a) weight = 452 × 9.81
component down the slope = 452 × 9.81 × sin 14°
= 1072.7 = 1070 N
M1
A0
[1]
C1
C1
A1
[3]
C1
A1
[2]
v = (0 + 2 × 0.13 × 10)1/2 = 1.61 or 1.6 m s–1
A1
[1]
(c) straight line from the origin
line down to zero velocity in short time compared to stage 1
line less steep negative gradient
final velocity larger than final velocity in the first part – at least 2×
B1
B1
B1
B1
[4]
(b)
(i)
(ii)
w
w
F = ma
T – (1070 + 525) = 452 × 0.13
T = 1650 (1653.76) N any forces missing 1/3
1.
/: /
tt p
2.
2
s = ut + ½at hence 10 = 0 + ½ × 0.13t
t = [(2 × 10) / 0.13]1/2 = 12.4 or 12 s
2
h
GCE AS/A LEVEL – October/November 2012
2
C1
A1
21
s
y
0.2, 0.6, 1.0 s (one of these)
[1]
9702
22
(a) (resultant) force = rate of change of momentum / allow proportional to
or change in momentum / time (taken)
(b)
(i)
∆p = (–) 65 × 10–3 (5.2 + 3.7)
= (–) 0.58 N s
(ii)
C1
[2]
A1
[1]
M1
A1
[2]
B1
[1]
B1
kinetic energy (of ball and wall) is reduced / not conserved so inelastic
(Allow relative speed of approach does not equal relative speed of separation.)
[1]
F = 0.58 / 7.5 × 10–3
1.
2.
(ii)
[1]
A1
= 77(.3) N
(c) (i)
B1
force on the wall from the ball is equal to the force on ball from the wall
but in the opposite direction
(statement of Newton’s third law can score one mark)
momentum change of ball is equal and opposite to momentum change
of the wall / change of momentum of ball and wall is zero
[email protected] Cell No. 03467030039
(a) Resultant force (and resultant torque) is zero
Weight (down) = force from/due to spring (up)
i
w
s
ic
GCE AS/A LEVEL – May/June 2012
3
A1
m
o
1280 N
(i)
41
23
weight = m × g
= 130.5 × 9.81 = 1280 N
(ii)
(c)
9702
42
GCE AS/A LEVEL – May/June 2013
3
(a)
(b)
(i)
9702
the total momentum of a system (of interacting bodies) remains constant
provided there are no resultant external forces / isolated system
M1
A1
[2]
(ii)
elastic: total kinetic energy is conserved, inelastic: loss of kinetic energy
[allow elastic: relative speed of approach equals relative speed of separation]
B1
[1]
(i)
initial mom: 4.2 × 3.6 – 1.2 × 1.5
final mom: 4.2 × v + 1.5 × 3
v = (13.3 – 4.5) / 4.2 = 2.1 m s–1
C1
C1
A1
[3]
M1
M1
A1
[3]
(ii)
c
i
s
w
s
9702
(a) a body / mass / object continues (at rest or) at constant / uniform velocity unless
acted on by a resultant force
(b)
(i)
B1
[1]
B1
B1
[2]
acceleration = gradient or (v – u) / t or ∆v / t
= (6.0 – 0.8) / (2.0 – 0.0) = 2.6 m s–2
C1
M1
[2]
F = ma
= 65 × 2.6
= 169 N (allow to 2 or 3 s.f.)
A1
[1]
C1
C1
A1
[3]
2.
3.
p
.
w
w
w
weight component seen: mg sinθ (218 N)
218 – R = 169
R = 49 N
(require 2 s.f.)
/: /
tt p
h
y
h
weight vertically down
normal / reaction / contact (force) perpendicular / normal to the slope
(ii) 1.
22
[email protected] Cell No. 03467030039
c
.
it
= ½ mA(vA)2 + ½ mB(vB)2
= 27.21 + 1.08 = 28(.28)
final kinetic energy = 9.26 + 6.75 = 16
initial KE is not the same as final KE hence inelastic
provided final KE less than initial KE
[allow in terms of relative speeds of approach and separation]
initial kinetic energy
/
m
o
(= 15.12 – 1.8 = 13.3)
GCE AS/A LEVEL – May/June 2014
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