Teachers’ Copy
2009 Physics JC1 H1 & H2
Name: …………………………………………….…
CG : ……………
Date: __ Feb 09
Tutorial 2 : Kinematics
Complete Q1- 3 with reference to lecture notes Eg. 1-8.
2(a) An object is travelling along a straight line. The
graph shows the variation with time of its
position during the whole journey.
1(a) Can an object have zero velocity and still be
accelerating?
Yes. Acceleration is the rate of change of
velocity. It causes velocity to change. When
the direction of velocity and accelerating of
an object are in the opposite direction, the
object will slow down till its velocity
become zero and subsequently speed up in
the same direction of the acceleration (i.e.
change direction of motion).
(b)
Identify the position/s where the object is
(i) moving fastest, D (greatest speed)
(ii) at rest,
B, E (zero speed)
(iii) slowing down A (speed becoming smaller)
(iv) changing direction. B (change in sign of
Can an object have constant velocity and still
have varying speed? How about the reverse?
velocity)
No. Constant velocity means the magnitude
(i.e. speed) and direction are fixed.
Yes. It is possible to have constant speed
while the direction varies eg circular
motion.
(c)
(b)
Two objects C and D are travelling along a
straight line. The graphs show their variation
with time of their positions during the whole
journey. Do objects C and D ever have the
same velocity?
Can the velocity of an object reverse direction
when the object’s acceleration is constant?
Yes, see part (a)
(d)
Can an object be increasing in speed as its
acceleration decrease?
Yes. Acceleration is the rate of change of
velocity, thus decreasing acceleration will
indicate decreasing rate of change of
acceleration. Thus the object’s velocity is
increase but at a decreasing rate, i.e. speed
of object is increasing but at a decreasing
rate.
Decreasing acceleration is not deceleration.
(e)
Yes, velocity is the rate of change of
displacement (indicated by the gradient of a
displacement-time
graph).For
same
velocity, the gradient of a point on graph C
and D should be the same. The graphs of
objects C and D have the same gradient at
time 2.0 s.
(the intersection of C and D indicates same
displacement at that time instant, not same
velocity.)
Can the instantaneous velocity of an object at an
instant of time ever be greater in magnitude than
the average velocity over a time interval
containing that instant? Can it ever be less?
Yes. Yes. The average velocity must be
between the largest and smallest values.
(f)
3.
Can the average speed be larger than the
greatest instantaneous speed during that time
interval?
A second ball is dropped from a cliff 1.0 s after a
first ball was dropped. As both fall, does the
distance between them increase, decrease, or
stay the same?
Since the first ball is dropped earlier than
the second ball, it has accelerated earlier
than the second ball. Thus its velocity is
always greater than the second ball. With a
greater velocity, the first ball will
experience
a
greater
change
in
No. The average value cannot be
greater than the greatest value.
Eg Average height of the students in a
class cannot be greater than the ht. of
the tallest student.
1
displacement. Thus the distance between
the two balls should increase.
Mathematical Proof:
s1 = ½ g t2 s2 = ½ g(t-1)2
Separation = s2 - s1 = ½ g(t-1)2 - ½ g t2
= g [t- ½ ]
Complete Q4-7 with reference to lecture notes Eg 9-11.
4. Consider the following combinations of signs and
values for the velocity and acceleration of a car
with respect to a one-dimensional x-axis.
Velocity
Acc.
Describe outcome
(a)
positive
positive
Car speeds up in
+x direction.
(b)
positive
negative
Car slows down
in +x direction.
(c)
positive
zero
Car is moving
with constant
velocity in +x
direction.
(d)
negative
positive
Car slows down
in –x direction.
(e)
negative
negative
Car speeds up in
–x direction
(f)
negative
zero
Car is moving
with constant
velocity in –x
direction.
(g)
zero
positive
Car accelerates
from rest in +x
direction.
(h)
zero
negative
Car accelerates
from rest in –x
direction.
Rectilinear Motion
6. N96/I/4
A car is travelling along a straight road. The graph
shows the variation with time of its acceleration
during part of the journey. At which point on the
graph does the car have its greatest velocity?
Describe what the car is doing in each case.
Assume rightward direction is positive.
5.
Ans : C
Acceleration is the rate of change of velocity.
Thus, having a positive acceleration, the velocity
will be increasing. For greatest velocity, the car
should be accelerating. From 0 to C, the car is
having positive acceleration, i.e. the car’s
velocity is increasing. After point C, its
acceleration is negative, i.e. velocity is
decreasing. Thus the velocity of the car is
greatest at point C.
Ans : A
Acceleration is the rate of change of
velocity (indicated by the gradient of the
velocity-time graph).
(Alternatively, greatest change in velocity is also
indicated by positive area under velocity-time
2
A student wishes to measure the length of a
metal plate. The only equipment available is an
electronic timer controlled by a light beam and a
rod 1.00 m long. Using the rod, the student
positions the plate so that its lower edge is
1.00 m above the light beam.
graph since area under velocity-time graph
indicated the change in velocity)
7. N97/II/1 (modified)
The figure shows a velocity-time graph for a
journey lasting 65.0 s. It has been divided up into
six sections for ease of reference.
The metal plate is released and the timer starts to
record as the light beam is cut. The total time for
the plate to pass through the beam is 0.052s. The
student is told that the local value of the
acceleration of free fall is 9.79 m s-2.
(a)(i) Calculate the time for the bottom edge of
plate to reach the light beam.
[0.452 s]
Taking upwards as positive,
s = ut +
(a) Using information from the graph, obtain
i. the velocity 5.0 s after the start,
[10 ms-1]
ii. the acceleration in sect. A,
[2.0 m s-2]
iii. the acceleration in sect. B,
[0 m s-2]
iv. the acceleration in sect. E,
[- 7.0 m s-2]
v. the displacement travelled in sect. B, [300 m]
vi. the displacement travelled in sect. C. [250 m]
1
1.00  0t  (9.79)t 2
2
t = 0.452 s
(ii) Calculate the length of the metal plate, giving
your answer to an appropriate number of
significant figures.
[0.243 m]
(b) Describe qualitatively in words what happens in
sections E and F of the journey.
s2 = ut+
In E: The object experiences a constant
negative acceleration -7.0 m s-2 where its
velocity change from + 30 m s-1 to – 5 m s-1.
Note: when velocity of object is 0, it is at a
turning point (i.e. maximum displacement).
The object then speeds up to 5 m s-1 in the
reverse direction.
In F: The object maintains a constant speed
of 5 m s-1 in the reverse direction.
B
C
D
E
1 2
at
2
1
(1.00  x)  0  (0.452  0.052)  (9.79)(0.452  0.052)2
2
x = 0.243 m
(b)
1.
(c) Sketch the shape of the corresponding
displacement-time graph. You are not expected
to make detailed calculations of the displacement
traveled.
A
1 2
at
2
2.
3.
F
9.
(c)
Complete Q8-9 with reference to lecture notes Eg 12-17.
8.
3
Suggest two reasons why the time for the
bottom edge of the plate to reach the light
beam may differ from that quoted in (a)(i).
The presence of air resistance would
retard the motion which affect the time of
flight of the falling metal plate.
Miscalibration of stopwatch.
Zero error in stopwatch.
N03/III/1 (part)
The graph below shows the variation with time t
of the velocity v of the ball from the moment it is
thrown with a velocity of 26 m s -1 vertically
upwards.
(i)
State the time at which the ball reaches its
maximum height.
t = 1.80 s (when vy = 0)
(ii)
State the feature of a velocity time graph that
helps determine its acceleration.
Slope of the graph gives the instantaneous
acceleration.
(iii) Just after the ball leaves the thrower’s hand, it
has a downward acceleration of approximately
20 m s-2. Explain how this is possible.
(Acceleration is affected by resultant force.
In free-fall, the only force acting on the
object is weight, which accounts for an
acceleration of 9.81 m s-2). To have an
acceleration of 20 m s-2, the resultant force
acting on the ball must be greater than
weight. The additional force can be
accounted by the air resistance acting on
the ball. When a ball is thrown up, there is
a downward air resistance acting on it. The
air resistance plus the weight of the ball
cause the (downward) resultant force on
the ball to be increased.
Fn = ma
Fair + W = ma
(Note: Air resistance is dependent on the
speed of the ball).
(d)
Motion Up
W
R
The weight and air
resistance both act
downwards. Greater
net opposing force.
Thus greater
downwards
acceleration. Shorter
time to cover the same
distance (come to rest
at maximum height)
t = 1.80 s. This is when the ball is
momentarily at rest and hence air
resistance does not act. (Fair = 0 when v =
0)
W
The air resistance acts
opposite to the weight.
Smaller net force is
acting. Thus smaller
downward
acceleration. Longer
time to cover the same
distance to reach
bottom (starting from
rest).
Complete Q10-12 with reference to lecture notes Eg 18-19.
Projectile Motion
10. J04/I/9
A motorcycle stunt-rider moving horizontally
takes off from a point 1.25 m above the ground,
landing 10 m away as shown.
(v) Sketch an acceleration time graph for the motion.
Show the value of g on the acceleration axis.
a
Motion down
R
(iv) State the time at which the acceleration is g.
Explain why the acceleration has this value only
at this particular time.
0
Explain why, for all real vertical throws, the time
taken to reach maximum height must be shorter
than the time taken to return to the starting point.
1.80 s
t
g = -9.81 m s-2
Gradient of
v-t graph
What was the speed at take-off?
Note : The graphs below contrast that between
with and without air resistance.
Taking upwards as positive,
sy  u y t 
1 2
ay t
2
1
 9.81 t 2
2
t = 0.5048 s
1.25  0t 
4
[19.8 m s-1]
sx  u x t 
1 2
ax t
2
12. In a certain crash test conducted by ‘Q’ on the
latest gadget-filled car built especially for 007
Mr. James Bond, a convertible was driven off the
cliff with an initial speed of 20.0 m s-1 at a tilted
angle of 30o as shown below.
1
2
00.5048 
2
ux = 19.8 m s-1
10  ux  0.5048  
o
11. A football is kicked at an angle of 37.0 with a
velocity of 20.0 m s-1. Calculate
(a)
(b)
(c)
(d)
(e)
the maximum height
[7.38 m]
the time of travel before the football hits the
ground
[2.45 s]
how far away it hits the ground
[39.1 m]
the velocity at the maximum height.
[16.0 ms-1]
the acceleration at the maximum height.
[9.81 ms-2]
(a) the maximum height
ux = 20 cos(370) = 16.0 m/s
uy = 20 sin(370) = 12.0 m/s
A crash dummy was ejected with a speed of
5.0 m s-1 in a direction perpendicular to the car
at the precise moment when the car reached the
edge of the cliff which is 50.0 m from the ground.
Treat car and dummy as point masses. Assume
negligible air resistance.
Vertically: vy2 = uy2 + 2aysy
0 = 122 + 2(-9.81)sy
sy = 7.38 m
(b) the time of travel before the football hits
the ground
Preliminary step
ux
vertically: vy = uy + ayt
-12 = 12 + (-9.81)t
t = 2.45 s
Let u = initial speed of car
(c) how far away when it hits the ground
ux = u cos = 20.0 cos 30
o
uy = u sin = - 20.0 sin30
30o
u = 20.0 m s-1
o
range = ux t = 16.0  2.45 = 39.1 m
(a) Determine the time taken for the car to hit the
ground.
[2.33 s]
(d) the velocity at the maximum height.
Since vy = 0, v = vx only
v at max height = 16.0 m s-1 →
sy = uyt + ½ ay t2
o
-50.0 = (-20.0 sin 30 ) (t) + ½ (-9.81)(t2)
o
(4.905) t2 + (20.0 sin 30 ) t – 50.0 = 0
t = 2.332 = 2.33 s
(e) the acceleration at the maximum height.
Since ax = 0, a = ay only
accel at max height = 9.81 m s -2 ↓
(b) Determine the velocity of the car just before it hits
o
the ground. [37.2 m s-1; 62.2 below the horizontal line]
In x-direction
o
vx = ux = 20.0 cos30 = 17.32 m s-1
In y-direction
vy = uy + ay t
o
= (- 20.0 sin30 ) + (-9.81)(2.332)
-1
= -32.88 m s
5
uy
Alternatively
vy2 = uy2 + 2ays
o
vy2 = (- 20.0 sin30 )2 + 2(-9.81)(-50.0)
= 1081
vy = -32.88 m s-1
(e) Determine the height of the crash dummy relative
to the ground at the moment the car hits the
ground.
[30.3 m]
sy = uyt + ½ ay t2
o
= (+25.0 sin 6.87 ) (2.332)
+ ½ (-9.81)(2.3322)
sy = -19.70 m
vx
v  17.32 2  32.88 2

= 37.2 m s-1
vy
v
tan 
vy
vx

32.88
 1.898
17.32
 = 62.22 = 62.2
o
Height of dummy (H) rel. to ground
= 50.0 – 19.7
H = 30.3 m
o
The car crashes with a speed of 37.2 m s-1
o
at an angle of 62.2 below the horizontal.
(c)
(f)
sy = uyt + ½ ay t2
o
-50.0 = (+ 25.0 sin 6.87 ) (t) + ½ (-9.81)(t2)
(4.905) t2 – 2.99 t – 50.0 = 0
t = 3.512 = 3.51 s
Determine the distance between the point where
the car hits the ground and the side of the cliff.
[40.4 m]
sx = ux t
o
= (20.0 cos30 )(2.332)
= 40.39 = 40.4 m
(d)
Determine the time taken for the dummy to hit
the ground.
[3.51 s]
(g)
Determine the initial velocity of the just-ejected
dummy.
[25 m s-1; 6.9° above horizontal]
Determine the distance between the point where
the dummy hits the ground and the side of the
cliff.
[87.1 m]
sx = ux t
o
= (25.0 cos6.87 )(3.512)
= 87.12 = 87.1 m
Note the dummy has two components of
velocities (parallel and perpendicular).
Formula
Acceleration due to free fall = 9.81 m s-2
u  15.0 2  20.0 2
= 25.0 m s
v = u + at
-1
s=
15.0
 0.750
20.0
o
o
 = 36.87 = 36.87
1
2
u + v  t
s = ut + 21 at 2
tan 
v 2 = u 2 + 2as
Angle of projection (elevation)
o
o
= 36.87 -30.0
o
= 6.87
The initial velocity of the dummy is 25.0 m
o
s-1 at an angle of 6.87 from the horizontal.
6
2009 Physics JC1 H1 & H2
20
Name: …………………………………………….(
)
CG : ……………
Date: ……..
Assignment 2 : Kinematics
Deadline for submission : ………………
Write your answers on a foolscap paper. Attached it to this question paper and give it to your
Physics Rep. who will then collect all the assignments and submit to your Physics Tutor by the
above deadline.
Assume acceleration due to free fall = 9.81 m s-2
1(a)
(b)
(i) Distinguish between speed and velocity.
[1]
(ii) Define acceleration.
[1]
A hot-air balloonist drops an apple over the side while the balloon is accelerating
upward at 4.0 m s-2 during lift-off.
(i) What is the apple’s acceleration once it has been released?
[1]
-1
(ii) If the velocity of the balloon is 2.0 m s upward at the instant of release, what is
the apple’s velocity just then?
[1]
(iii) At the point of release, the apple is 100 m above the ground. Calculate the time it
takes before it hits the ground.
[4.72 s]
[1]
(iv) A second apple is projected horizontally with a speed of 5.0 m s-1 at the precise
moment the first apple was released from the height of 100 m. Calculate the
horizontal distance travelled by the second apple.
[23.6 m]
[1]
(v) Determine the velocity of the second apple just before it hits the ground.
[3]
o
[44.6 m s-1, at 83.6 below the horizon line]
(v) Sketch the paths taken by the first and second apples.
7
[2]
2.
William Tell, an expert marksman, was punished when he did not bow before the evil
emperor. To receive his pardon, he was forced to shoot an apple off the head of his son,
Walter, or else both would be executed.
The apple of diameter 0.10 m was placed on top of his son’s head. The centre of the apple
is placed such that it is exactly along the horizontal line of the arrow. The horizontal line
is 1.50 m above the ground. William Tell is to shoot at the distance of 25.0 m. The initial
speed of the just released arrow is 90.0 m s-1.
Figure is not drawn to scale.
(a) If William Tell were to shoot the arrow horizontally, he is not likely to be successful
in hitting the apple. Why?
[1]
(b) Determine the drop from the horizontal line had he shot the arrow horizontally. [2]
(c) In order to hit the apple squarely in the centre, he needs to aim the arrow higher.
Determine the angle of projection.
[4]
(d) State two assumptions made when obtaining the angle in part (c).
End
8
[2]
Assignment 2 (Solution) Kinematics
1(a) (i) Speed is a scalar quantity while velocity is a vector quantity.
Speed has magnitude only while velocity is defined by its magnitude and direction.
(ii) Acceleration is the rate of change of velocity.
[1]
[1]
(b) (i) 9.81 m s-2 downwards (affected by resultant force (i.e. weight) acting at that time instant) [1]
(ii) 2.0 m s-1 upwards (same velocity as the moment just before the release (inertia))
(iii)
sy = uyt + ½ ayt2
-100 = (+2.0)t + ½ (-9.81)t2
2
4.91t -2t – 100 = 0
t = 4.72 s
[1]
[1]
(iv) Since the second apple has the same vertical velocity of + 2.0 m s-1 as the first apple, it has
the same flight time as the first apple, i.e. t = 4.72 s
Horizontal distance = ux t = (5.0) (4.72) = 23.6 s
[1]
(v)
In the y-direction
In the x-direction
vy = uy + ay t
= + 2.0 + (- 9.81)(4.721)
= - 44.3 m s-1
vx = ux = + 5.0 m s-1
[1]
Resultant Velocity =
tan  
vy
vx

44.3  5.0
44.3
 8.86
5.0
2
2
= 44.6 m s
 = 83.6
-1
[1]
o
[1]
(vi)
y
Note: Both apples has an
initial upwards velocity due to
its inertia. Thus it should be
going up initially till its vy = 0
1st Apple [1]
2nd Apple [1]
x
9
2 (a) Because after the arrow is released, it will drop due to gravity pulling it down while moving
forward. It is only in the absence of gravity, will it travel in a straight line.
[1]
(b)
Consider the x-direction
Time taken to travel to target = 25.0 / 90.0 = 0.278 s.
[1]
Consider the y-direction
Drop = sy = ½ gt2 = ½ (9.81)(0.278)2 = 0.378 m
[1]
Since the drop is greater than the radius of the apple (0.05 m), the arrow does not hit the
apple..Ouch!
(c) Let the angle of projection = 
sx = uxt + ½ ax t2
25.0 = 90.0 cosθ t + ½ (0)t2
25.0
Time of flight, T =
90.0 cos
[1]
Consider the y-direction
sy = uy t + ½ ayt2
0 = 90.0 sin (
25.0
25.0
) - ½ (9.81)(
)2
90.0 cos
90.0 cos
2sincos = sin 2 = 0.0302
2 = 1.735
o
 = 0.868
[1]
[1]
[1]
Angle of projection is 0.868o above the horizontal line.
2
*Shortcut …. Use directly R = u sin 2 (only for sy = 0 when the initial and final point are in
g
the same vertical height)
(d) Assume there is no air resistance
and acceleration due to free fall is constant.
(alternative: The arrow and apple are point objects.)
10
[1]
[1]