2016 j1 h2 kinematics tutorial guide 2016-2.pdf

8/19/2019 2016 J1 H2 Kinematics Tutorial Guide 2016-2.pdf

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Anglo-Chinese Junior College Tutorial SolutionTutorial: Kinematics H2(9749)

JC1 2016 of 6

Tutorial 2: Kinematics (Guide)

Discussions

1 (a)(i) Yes. When a body is thrown vertically upwards, It is instantaneously at restat the top of its trajectory. It is accelerating (downwards) becausegravitational force is acting on the body.[At the next instant, there is an increase in the body’s velocity]

(ii) Yes. When a body is moving with constant velocity.

(iii) Yes. When a body is slowing down during its motion implies it isundergoing deceleration. i.e. v and a are in opp direction.

(b) The displacement of an object measures the change in position of theobject along a straight line and is defined as the distance and direction of aposition of a point as measured from a reference position.

Hence if the car returns back to its initial position, its displacement is zerobut the total distance travelled is not zero.

(c) Speed is a physical quantity and hence must be defined in terms ofphysical quantities. The second used in the statement is the SI unit fortime. Hence the correct statement should be speed is the total distancetravelled per unit time.

2 D

3(a) B

(b) Graph A – straight line through origin and positive gradientGraph B – horizontal line (negative value)

Graph C – zero velocity (stationary)

Graph D – straight line with negative gradient

4(a)

a

s

v

t

t

t

u

The initial gradient of the

s-t graph is not zero.

0 0

0




JC1 2016 of 6

4(b)

Questions : Displacement – time graph

5(a) D (Answers can be seen from a- t graphs)

(b)

avt

t

u

t

0

0s

0

x x

t tt

x x

t

v v

t

v

t

v

tt

a a

t

a

t

a

t

t




JC1 2016 of 6

6(a) v is -ve for 0 s < t < 3 s , as slope of x-t graph is negative.

v is +ve for 3 s < t 7 s , as slope of x-t graph is positive.

v is zero for t = 3 s , as slope of x-t graph is zero.

(b) The velocity of an object at time t is obtained from the gradient of the tangent drawn atthe displacement – time graph at t .

(c)

Questions : velocity – time graph

7 For questions (a), (b), (c), apply the concept – distance travelled is given

by the area under a velocity-time graph

(a) 105.6 m (106 m)

(b) 41.7 m

(c) 492.06 m (492 m)

8(a)(i) A: a = (20 - 0)/10 = 2.0 m s-2 (Full workings must be shown; a = 20/10 not

acceptable)

E: a = (-5 – 30)/5 = - 7.0 m s-2 (Full workings must be shown; a = -35/5 not

acceptable)

(ii) Using area under the graph

B: distance = 20 x 15 = 300 m

C: distance = ½ (20 + 30)(10) = 250 m

(b) Object decelerates uniformly from 30 m s-1 to zero in about 54 s. After

which it reverses its direction of travel and accelerates to 5 m s-1 in the

opposite direction in about 1.0 s. It then travels with constant velocity of 5

m s-1 in the opposite direction to its initial direction of travel in section F.

2

30

4 6

v

t / s




JC1 2016 of 6

(c)

Questions : acceleration – time graph

9 C

Questions : Motion of object under uniform gravitational field

10 (a) S - R

(b) R

(c) R + S

11 At time tx , a = ½ g tx 2

At time ty , a + h = ½ g ty 2

½ g tx 2 + h = ½ g ty 2

g = 2 h / (ty 2 - tx 2)

Questions : Motion of object under uniform gravitational field and air resistance

12(a) 1.75 s or 1.80 s

(b) The magnitude of the gradient of the tangent at the given time gives the

magnitude of the acceleration at that time.

(c) Acceptable value of ai is 19.8 to 20.2 m s-2.

(d) Acceptable value of av=0 is 9.5 to 10.5 m s-2.

(e) This value is close to 9.81 m s-2 which is acceleration of free fall (

acceleration due to gravity without air resistance)

Distance -

time

Displacement -

time

Highlight that graph must be

continuous

x/m

t/s0 10 20 30 40 50 60 70




JC1 2016 of 6

(f)

13 C

14(a)(i) Taking downwards as +ve

sy = uy t + ½ ay t2

1.20 = 0 + ½ 9.81 t2

t = 0.495 s

(ii) sx = ux t

1.52 = ux 0.495 ux = 3.07 m s-1

(iii) vx = ux = 3.07 m s-1

vy = uy + ay t = 0 + 9.81 (0.495) = 4.86 m s-1

(iv)Magnitude of velocity = 75.586.407.3

22 m s-1

o

7.57

07.3

87.4tan

and

o

7.57 below the horizontal

(b) No change from (a)(i) since for both cases, the initial vertical velocity arezero and that they both experience the same acceleration due to gravity.

15(a) m9.6745cos96 o

x s

(b)1-

sm8.15s4.3

m9.67

x u

(c) m9.6745sin96 o

x s

vx

Vy

VR




JC1 2016 of 6

15(d)

1-

2

2

sm3.5

3.481.92

1)3.4(9.67

2

1

y

y

y y

u

u

at t us

(e) (f) Magnitude of velocity = 7.168.153.5 22

m s-1

o

5.18

8.15

3.5tan

ando

5.18 above the horizontal

16 C

17 (a) At the top of its trajectory, the speed of the ball is a minimum as vy = 0,

(b) hence speed of ball = vx = 19 m s-1

(c) sx = uy t = 19 (5) = 95 m

(d) At time t = 0 s, u = 31 m s-1

u = ( ux 2 + uy 2 )1/2

31 = ( 19 2 + uy 2 )1/2 uy = 24.5 m s-1

(e) Taking upwards as +ve

sy = uy t + ½ ay t2 = 24.5 (2.5) - ½ 9.81 (2.5)2 = 30.6 m

uy uR

UX

2016 j1 h2 kinematics tutorial guide 2016-2.pdf

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