{{<P=21><C=06><S=extension><T=SQ><M=6><L=1><X=H><id=001>}}

2162001

E An aid parcel is released from a plane flying horizontally at 60 m s–1 at an altitude of 1 km. Take the acceleration due to gravity as 10 m s−2.

(a) What are the initial horizontal and vertical velocities of the parcel? (2 marks)

(b) How long does the parcel take to hit the ground? (Neglect air resistance.)

(2 marks)

(c) At what horizontal distance should the plane be from the target when the parcel is released? (2 marks)

--ans --

Solution

Marks

(a) Horizontal component: ux = 60 m s–1

Vertical component: uy = 0 m s–1

1A

1A

(b) Take the downward direction as positive.

s =

1000 =

t = 14.14 s

≈ 14.1 s

1M

1A

(c) Horizontal distance = uxt

= 60 × 14.14

= 848 m 

1M

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=5><L=1><X=H><id=002>}}

2162002

E (a) What is the maximum range possible for a projectile fired from a cannon if the projectile is fired at a speed of 150 m s–1? (2 marks)

(b) What is the maximum height reached in (a)? (2 marks)

Take the acceleration due to gravity as 10 m s−2.

-- ans--

Solution

Marks

(a) Maximum range =

1M

=

= 2250 m

1A

(b) Maximum height =

1M

=

= 563 m

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=8><L=2><X=H><id=003>}}

2162003

★

E A stationary trolley of mass 2 kg is placed at the edge of a horizontal smooth platform as shown. The platform is 0.2 m above the ground. A small bearing of mass 0.1 kg is projected from the ground with a velocity of 4 m s–1 at an angle 30°. The bearing hits the trolley, rebounds and eventually falls onto the ground at a horizontal distance 0.4 m from the edge of the platform.

Take the acceleration due to gravity as 10 m s−2. Neglect the air resistance and frictions. 

(a) Find the velocity of the bearing when it hits the trolley. (3 marks)

(b) Find the speed of the trolley after it is hit by the bearing. (5 marks)

-- ans --

Solution

Marks

(a) Consider the vertical motion of the bearing (taking upwards as positive):

By v2 = u2 + 2as,

1M

vy == 0

1M

Velocity of the bearing 

= vx = 4 cos 30° = 3.46 m s–1 (horizontally to the right)

1A

(b) Let the time of flight after rebound be t.

Consider the vertical motion of the bearing when it rebounds (taking the downward direction as positive):

By s = ut +at 2,

0.2 = 0 +(10) t 2

t = 0.2 s

Consider the horizontal motion of the bearing when it rebounds (taking the direction to the right as positive):

The horizontal velocity of the bearing after rebound

=== –2 m s−1

By conservation of momentum,

mBuB + mTuT = mBvB + mTvT

0.1 × 4 cos 30° = 0.1 × (−2) + 2vT

vT = 0.273 m s–1 

∴ The speed of trolley is 0.273 m s–1.

1M

1M

1M

1M

1A

-- ans end --

{{<P=21><C=06><S=extension><T=SQ><M=6><L=2><X=H><id=004>}}

2162004

★

E Michele is going to strike a golf ball of mass 45.9 g into a hole 150 m away from her. She uses an iron club to send the ball at an angle of 30° to the horizontal. Assume that Michele and the hole are on the same level. Neglect air resistance and take the acceleration due to gravity as 10 m s−2.

(a) Find the speed of the ball when it leaves the club face. (2 marks)

(b) If the time of impact between the club face and the ball is 0.05 s, what should be the average force applied on the ball during the strike in order to achieve the initial speed calculated in (a)? (2 marks)

(c) Sketch the trajectories of the ball on the same graph when 

(i) effects of air resistance are negligible, and (1 mark)

(ii) effects of air resistance are taken into account. (1 mark)

-- ans --

Solution

Marks

(a) By range =,

1M

150 =

u = 41.6 m s−1

∴ The speed of the ball is 41.6 m s–1.

1A

(b) By Ft = mv − mu,

1M

average force === 38.2 N

1A

(c)

(symmetrical parabolic curve for no air resistance) 

(asymmetric curve showing shorter range and reduced maximum height in the presence of air resistance) 

1A

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=7><L=2><X=H><id=005>}}

2162005

★

E A bullet is fired horizontally from a gun that is 50 m above the ground. The initial speed of the bullet is 350 m s–1. Take the acceleration due to gravity as
10 m s−2.

(a) What is the time of flight of the bullet in the air? (2 marks)

(b) What is the horizontal distance between the firing point and the position that the bullet strikes the ground? (2 marks)

(c) What is the speed of the bullet when it hits the ground? (3 marks)

-- ans --

Solution

Marks

(a) In the vertical direction:

The initial vertical velocity is zero.

1M

1A

(b) In the horizontal direction:

Distance = ut 

= 350 x 3.16

= 1110 m

1M

1A

(c) When the bullet hits the ground,

m s–1

350 m s–1

1M

1M

Speed 351 m s–1

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=9><L=2><X=H><id=006>}}

2162006

★

E A 0.15-kg ball leaves an edge at 2.4 m s–1 as shown in the figure. Assume air resistance is negligible. Take the acceleration due to gravity as 10 m s−2.

(a) What is the vertical component of the ball’s momentum when the ball strikes the floor 0.85 m below? (4 marks)

(b) What is the velocity of the ball when it strikes the floor? (2 marks)

(c) Calculate the time of flight of the ball. (2 marks)

(d) Find the horizontal distance travelled by the ball before it strikes the floor. (1 mark)

-- ans --

Solution

Marks

(a) In the vertical direction (taking the downward direction as positive):

By v2 = u2 + 2as,

1M

v ==== 4.12 m s−1

1A

∴ Vertical component of the ball’s momentum = mv

= 0.15 ×

= 0.618 kg m s–1

1M

1A

(b) Speed of the ball === 4.77 m s–1

1M

The direction is given by:

tan θ ==

θ = 59.8° 

∴ The velocity of the ball is 4.77 m s–1 at an angle of 59.8° below the horizontal.

1A

(c) In the vertical direction (taking the downward direction as positive):

By v = u + at,

1M

time of flight === 0.412 s

1A

(d) Distance travelled = vt = 2.4 ×= 0.990 m

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=8><L=2><X=H><id=007>}}

2162007

★

E A ball is thrown from the ground with an initial velocity of 15 m s–1 at an angle of 60° to the ground. Take the acceleration due to gravity as 10 m s−2.

(a) (i) Find the initial vertical velocity of the ball. (1 mark)

(ii) Find the initial horizontal velocity of the ball. (1 mark)

(b) Assuming that air resistance can be neglected, use your answers in (a) to determine:

(i) the maximum height that the ball can reach; (2 marks)

(ii) the time of flight; (2 marks)

(iii) the range of the ball. (2 marks)

-- ans --

Solution

Marks

Take the upward direction as positive.

(a) (i) vy = 15 sin 60° = 13.0 m s–1

(ii) vx = 15 cos 60° = 7.5 m s–1

1A

1A

(b) (i) In the vertical direction:

By v2 = u2 + 2as,

0 = (15 sin 60°)2 + 2 (–10)s

s = 8.44 m

∴ The maximum height is 8.44 m.

(ii) In the vertical direction:

By v = u + at,

t === 1.30 s

Time of flight = 2 × 1.30 = 2.60 s

1M

1A

1M

1A

(iii) In the horizontal direction:

Range = vt

= 7.5 x 2.6

= 19.5 s

1M

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=6><L=2><X=H><id=008>}}

2162008

★

E A man throws a ball through a stationary hoop 5 m above his head as shown. The ball is travelling horizontally at 15 m s–1 when it passes through the hoop. Neglect air resistance and take the acceleration due to gravity as 10 m s−2.

(a) What is the initial vertical velocity of the ball? (2 marks)

(b) Find the time needed for the ball to travel to the hoop after leaving the man’s hand. (2 marks)

(c) Find the initial velocity of the ball. (2 marks)

-- ans --

Solution

Marks

Take the upward direction as positive.

(a) In the vertical direction:

When the ball passes through the hoop, its vertical velocity is zero.

By vy2 = uy2 + 2gs,

0 = u2y + 2(–10)(5)

uy = 10 m s–1

∴ The initial vertical velocity is 10 m s–1 upwards.

1M

1A

(b) In the vertical direction:

By vy = uy + at,

0 = 10 + (–10) t

t = 1 s

∴ The time needed is 1 s.

1M

1A

(c) Initial speed == 18.0 m s−1

1A

tan θ =

θ = 33.7°

∴ The initial velocity is 18.0 m s−1 at 33.7° above the horizontal.

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=6><L=2><X=H><id=009>}}

2162009

★

E A cannonball is fired with an initial horizontal speed of 50 m s–1 and vertical speed of 30 m s–1 at ground level. Assume air resistance is negligible and take the acceleration due to gravity as 10 m s−2.

(a) Find the equation of the trajectory of the cannonball. (3 marks)

(b) What is the range of the cannonball? (2 marks)

(c) What is the time of flight of the cannonball? (1 mark)

-- ans --

Solution

Marks

Take the upward direction as positive.

(a) The horizontal displacement is given by

x = uxt = 50 t 

1M

t = (1)

The vertical displacement is given by

y = uyt +a t 2

1M

y = 30t − 5t 2 (2)

Substitute (1) into (2),

y =

y = 0.6x – 0.002x2

1A

(b) When the cannonball reaches the ground, y = 0: 

0 = 0.6x – 0.002x2

x = 0  or  300 m

∴ The range of the cannonball is 300 m.

1M

1A

(c) From equation (1),

time of flight === 6 s

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=8><L=2><X=H><id=010>}}

2162010

★

E A stone is projected horizontally at 20 m s–1 from the top of a cliff 49 m above the sea. Take the acceleration due to gravity as 10 m s−2 and neglect air resistance.

(a) Find the time that it takes for the stone to reach the sea. (2 marks)

(b) Find the horizontal distance between the point where the stone hits the sea and the cliff. (1 mark)

(c) Find the speed of the stone when it hits the sea. (3 marks)

(d) If air resistance is taken into account, will the time of flight and the final velocity of the stone increase, decrease or remains unchanged? (2 marks)

-- ans--

Solution

Marks

(a) In vertical direction (taking the upward direction as positive):

By s = ut +,

1M

−49 = 0 +

t = 3.13 s

1A

∴ The time taken is 3.13 s.

(b) Horizontal distance = uxt = 20 × 3.13 = 62.6 m

1A

(c) In vertical direction (taking the downward direction as positive):

Vertical velocity = u + at 

= 0 + 10 × 3.13 

= 31.3 m s–1

1M

1M

Speed of the stone === 37.1 m s–1

1A

(d) Both will decrease.

2 × 1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=9><L=3><X=H><id=011>}}

2162011

★★

E As shown in the figure, a ball is projected horizontally with a speed of 8 m s–1 from point A on an inclined plane of inclination 45°. The ball then landed on another point B on the inclined plane. The acceleration due to gravity is 10 m s−2.

(a) Find the time of travel of the ball. (4 marks)

(b) Find the distance AB. (2 marks)

(c) Find the velocity of the ball at the instant before it lands. (3 marks)

-- ans--

Solution

Marks

Take the downward direction as positive.

(a) Horizontal displacement sx = uxt = 8t 

1M

Vertical displacement sy == 5t 2

1M

Since both A and B are on the inclined plane, 

tan 45° ==

1M

t = 1.6 s

1A

(b) Horizontal displacement sx = 8 × 1.6 = 12.8 m

1M

AB === 18.1 m

1A

(c) Horizontal velocity vx = 8 m s–1  

Vertical velocity vy = uy + at = 0 + 10 × 1.6 = 16 m s–1

1M

Speed of the ball === 17.9 m s–1

1A

tan θ ==

θ = 63.4°

∴ The velocity of the ball is 17.9 m s–1 at 63.4° below the horizontal.

1A

-- ans end--

{{<P=21><C=06><S=extension><T=SQ><M=9><L=3><X=H><id=012>}}

2162012

★★

E Roger strikes a tennis ball at C which is 2.10 m vertically above the edge A of the court. At the instant the ball is struck, it moves with a horizontal velocity v. As shown in the figure, it then just passes the net, hits the ground at D and reaches the highest point E. The court is 23.78 m long and the net is 0.914 m high.

Take the acceleration due to gravity as 10 m s−2 and neglect the effect of air resistance.

(a) Calculate the value of v. (3 marks)

(b) Calculate the speed of the tennis ball just before it hits the ground. (3 marks)

(c) If the collision between the ball and the ground is elastic, is E inside or outside the court? Explain your answer. (3 marks)

-- ans--

Solution

Marks

Take the downward direction as positive.

(a) Let t be the time required for the tennis ball to reach the net,

In the vertical direction:

By s =,

1M

2.10 − 0.914 = 0 +

t = 0.487 s

In the horizontal direction:

1M

v === 24.41 m s–1 ≈ 24.4 m s–1

1A

(b) In the vertical direction:

By vy2 = uy2 + 2as,

vy2 = 0 + 2 × 10 × 2.10 = 42

1M

1M

Speed of ball === 25.3 m s–1

1A

(c) Since the collision is elastic, E is at the same level as C. 

By symmetry, CD = DE.

Since AD is greater than 11.89 m,

E is outside the court.

1M

1M

1A

-- ans end--