Download MEI topic assessment Kinematics

EdExcel Mechanics 2
Kinematics of a particle
Chapter assessment
Take g = 9.8 ms-2 unless otherwise instructed.
1. A golf ball is hit over horizontal ground from a point O on the ground. The
velocity of projection is 30 ms-1 at 40° to the horizontal. The effects of air
resistance should be neglected.
(i) The ball is y m above the ground t seconds after projection. Write down an
expression for y in terms of t and hence determine the time at which the ball
first hits the ground.
[5]
The ball passes directly over a tree which is at a horizontal distance of 34 m from
O.
(ii) Determine the speed of the ball as it passes over the tree. Calculate also the
angle between the direction of motion of the ball and the horizontal at that
time, making it clear whether the ball is rising or falling.
[10]
2. In this question you should take g = 10 ms-2. The effects of air resistance
should be neglected.
A small stone is fired from a catapult 1 m above horizontal ground at a speed of
30 ms-1. The angle of projection with the horizontal is , where cos   0.6 and
sin   0.8 . The stone hits a vertical wall that is a horizontal distance of 27 m
from the point of projection. This information is shown in the diagram below
together with x- and y-axes and the origin O on the ground; the units of the axes
are metres.
y
wall
30 ms
Not to
scale
-1

1m
O
27 m
x
(i) Show that, after t seconds, the horizontal displacement of the stone from O,
x m, and the vertical displacement, y m, are given by
x  18t
and
y  1  24t  5t 2 .
(ii) What is the value of t when the stone hits the wall?
How high is the stone above the ground when it hits the wall?
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[3]
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EdExcel M2 Kinematics Assessment solutions
(iii)Show that the stone is rising when it hits the wall.
[2]
(iv) Find the horizontal displacement of the stone when it is at a height of 17 m
above the ground.
[5]
3. A particle moves on the x-axis. Its displacement, x m, from the origin O is given
by
x  3t 2  3t  2 , where t is the time in seconds.
How far is the particle from O when it is instantaneously at rest?
[5]
4. A racing car starts off down a straight section of track towards the first corner. Its
speed, v ms-1, is modelled for the first four seconds of its motion by
v  t 3  9t 2  24t ,
0t 4.
(i) Find an expression for the distance travelled by the car in the first t seconds.
Calculate the distance travelled from t = 2 to t = 4.
[5]
(ii) Show that the acceleration, a ms-2, of the car at time t is given by
a  k (t  2)(t  4) , where k is a constant to be determined.
[2]
5. The position vector, r, of a particle at time t is given by
r  t 2i  (5t  2t 2 ) j ,
where i and j are the standard unit vectors, lengths are in metres and time is in
seconds.
(i) Find an expression for the acceleration of the particle.
[4]
(ii) Is the particle ever at rest?
[2]
6. The velocity, v, of a particle is given as
v  2t 2  3t  13 t 3 .
(i) Show that the acceleration of the particle is zero when t = 1 and when t = 3.
[3]
(ii) Calculate the displacement of the particle from its position when t = 1 to its
position when t = 2.
[4]
7. An insect moves in a straight line. The time, t, is in seconds and distance travelled
is in metres.
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EdExcel M2 Kinematics Assessment solutions
The velocity, v ms-1, of the insect is given by
v  t 2  4t ,
0  t  6,
v  c,
6  t  10,
v  at  b,
10  t  15.
You are also given that v = 4 when t = 12.
(i) Show that c = 12.
[2]
(ii) Calculate the values of a and b and briefly describe the motion of the insect in
the interval 10  t  15 .
[4]
(iii) Calculate the values of v for t = 0, t = 2 and t = 4. Sketch the v-t curve for the
motion of the insect in the interval 0  t  6 .
[3]
(iv) Calculate the distance travelled by the insect in the interval 0  t  6 .
[6]
Total 70 marks
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EdExcel M2 Kinematics Assessment solutions
Kinematics of a particle
Solutions to Chapter assessment
1. (i) For vertical motion:
u  30 sin 40
sy
t t
g  9.8
s  ut  21 at 2
y  30t sin 40  4.9t 2
When the ball hits the ground, y = 0.
0  30t sin 40  4.9t 2
0  t (30 sin 40  4.9t )
30 sin 40
 3.94 (3 s.f.)
4.9
t = 0 is the time of projection,
so the time the ball first hits the ground is 3.94 seconds (3 s.f.)
t  0 or t 
(ii) Horizontal speed  30cos 40  22.9813
When it passes over the tree, 34  30t cos 40
t 
Vertically:
u  30 sin 40
34
30cos 40
v  u  at
v v
 30 sin 40 
34
30 cos 40
a  9.8
t 
 4.7849
9.8  34
30 cos 40
4.7849

22.9813
Speed  22.9813 2  4.78492
 23.5 (3 s.f.)
The speed of the ball as it passes over the tree is 23.5 ms-1 (3 s.f.)
4.7849
22.9813
  11.8 (to nearest degree)
tan  
The direction of motion of the ball is 11.8° above the horizontal.
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EdExcel M2 Kinematics Assessment solutions
The ball is rising, since its vertical velocity is positive.
2. (i) Horizontally, speed  30cos   30  0.6  18
Horizontal speed is constant, so x  18t .
Vertically, initial speed  30sin   30  0.8  24
Initial height = 1
u  24
s  ut  21 at 2
a  10
y  1  24t  5 t 2
s  y1
y  1  24t  5 t 2
t t
(ii) When the stone hits the wall, x = 27
x  18t  27  18t  t  1.5
When t = 1.5, y  1  24  1.5  5  1.5 2
 1  36  11.25
 25.75
(iii) When it hits the wall, vertically:
u  24
v  u  at
t  1.5
a  10
v ?
 24  10  1.5
 24  15
 9
Since the vertical speed is positive when it hits the wall, the stone is rising.
(iv) When y = 17, 17  1  24t  5 t 2
5 t 2  24t  16  0
(5 t  4)(t  4)  0
t  0.8 or t  4
Since the stone hits the wall at t = 1.5, the value of t when it is at a height
of 17 m is 0.8.
When t = 0.8, x  18t  18  0.8  14.4
The horizontal displacement of the stone is 14.4 m.
3. x  3t 2  3t  2
dx
 6t  3
dt
When particle is instantaneously at rest, 6t  3  0
t  0.5
When t = 0.5, x  3  0.5 2  3  0.5  2
 1.25
v 
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EdExcel M2 Kinematics Assessment solutions
It is 1.25 m from O when it is instantaneously at rest.
4. (i) v  t 3  9t 2  24t
s   v dt  41 t 4  3t 3  12t 2  c
When t = 0, s = 0  c = 0
s  41 t 4  3t 3  12t 2
Distance travelled from t = 2 to t = 4 is
 41  44  3  43  12  42    41  2 4  3  2 3  12  2 2 
 64  192  192  4  24  48
 36
Distance travelled = 36 m.
dv
 3t 2  18t  24  3 t 2  6t  8   3(t  2)(t  4)
dt
so k = 3.
(ii) a 
5. (i)
r  t 2 i  (5 t  2t 2 ) j
dr
 2t i  (5  4t ) j
dt
dv
a
 2i 4 j
dt
v
(ii) The velocity of the particle is given by 2t i (5  4t ) j . For the particle to
be at rest, both components of the velocity must be zero. The component of
the velocity in the i direction is only zero when t = 0, and the component
of the velocity in the j direction is only zero when t = 1.25. So the
particle is never at rest.
6. (i) v  2t 2  3t  31 t 3
dv
 4t  3  t 2
dt
When a = 0, t 2  4t  3  0
a
(t  1)(t  3)  0
t  1 or t  3
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(ii) Displacement   v dt
1
2
  23 t 3  23 t 2  121 t 4  1
  163  6  43    23  23 
1
12

13
  12
7. v  t 2  4t ,
v  c,
0  t  6,
6  t  10,
10  t  15.
v  at  b ,
(i) When t = 6, v  62  4  6  12
Therefore c = 12.
(ii) When t = 10, v = 12  10a  b  12
When t = 12, v = 4  12a  b  4
Subtracting: 2a  8  a  4, b  52
The insect is decelerating at a constant rate.
(iii)When t = 0, v  02  4  0  0
When t = 2, v  2 2  4  2  4
When t = 4, v  42  4  4  0
v
12
8
4
0
2
4
6
t
-4
4
4
6
6
(iv) Displacement for 0  t  4   v dt   31 t 3  2t 2  0
0
32
 64
3  32   3
Displacement for 4  t  6   v dt   31 t 3  2t 2  4
4
 72  72  64
3  32 
32
32
1
Total distance travelled = 3  3  21 3 m.
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