9702/Oct Nov/12/2017/Q7

The velocity-time graph for a train starting at one station and stopping at the next is shown.

Another train has double the acceleration but the same maximum speed and the same

deceleration.

Which velocity-time graph, on the same scale, shows the motion of this train between the same

stations?

Solution:

Answer: B

From option A to D, we know that all of the acceleration become double from beginning, after that same maximum speed and at the end same deceleration.

However, the question had mentions the same station for the train (which mean same displacement).

The displacement of the train = velocity x time
= the area of trapezium of the graph.
= 1/2 (11+5)(8)
= 64m

For option A, the displacement = 1/2 (9+5)(8) = 56m

For option B, the displacement = 1/2 (10+6)(8) = 64m

For option C, the displacement = 1/2 (8+4)(8) = 48m

For option D, the displacement = 1/2 (11+7)(8) = 72m

Reference: PYQ - Oct/Nov 2017 Paper 12 Q7

9702/Oct Nov/11/2015/Q7

One of the equations of uniformly accelerated motion is shown.

s = ut + ½ at²

Apparatus is arranged to record the time t taken for a marble to fall between two light gates

connected to timers. The marble touches the stop before it is released. The vertical distance s

between the light gates is measured.

Which graph does not show a correct relationship when light gate 2 moves up to light gate 1
which is fixed?

Solution:
Answer: C

Choice A: from the formula s = ut + ½ at² , we know s ∝ t²  , so choice A is correct.

Choice B: we can derive the formula of s/t = u + ½ at. s/t ∝ t , so choice B is correct.

Choice D: The gravitational acceleration is always constant. So choice D is correct.

Choice C:
We can derive the formula of
ut = s – ½ at²

u = s/t – ½ at

So u = 0 at all the time is doesn't make sense.

Reference: PYQ - Oct/Nov 2015 Paper 11 Q7

9702/May Jun/13/2015/Q18

A loaded aeroplane has a total mass of 1.2 ×10⁵ kg while climbing after take-off. It climbs at an angle of 23°to the horizontal with a speed of 50 m s^–1. What is the rate at which it is gaining potential energy at this time?

A.            2.3 × 10⁶ J s^–1

B.            2.5 × 10⁶ J s^–1

C.            2.3 × 10⁷ J s^–1

D.            2.5 × 10⁷ J s^–1

Solution:

Answer: C.

sin 23^o = y/50

y = 50 sin 23^o

    = 19.54 ms^-1

Use v² = u² + 2as to find the h

19.54² = 0 + 2 (9.81) S

S = 19.46 m

P.E. = mgh

        = (1.2 x 10⁵ )(9.81)(19.46)

         = 2.29 x 10⁷ Js^-1

Reference: PYQ - May/Jun 2015 Paper 13 Q18

9702/Oct Nov/11/2016/Q6

Question 6

A cyclist pedals along a raised horizontal track. At the end of the track, he travels horizontally into

the air and onto a track that is vertically 2.0 m lower.

The cyclist travels a horizontal distance of 6.0 m in the air. Air resistance is negligible.

What is the horizontal velocity v of the cyclist at the end of the higher track?

A      6.3 ms^–1 

B      9.4 ms^–1 

C      9.9 ms^–1 

D      15 ms^–1

Solution:

There are important points to note in this question:

The horizontal velocity v is used to calculate the horizontal distance

The time to reach the maximum height is the time to travel the horizontal distance

At maximum height u = 0

Using H = ut + 1/2gt^2

2 = 0 + 1/2×9.81xt^2

(t=0.6395s)

Horizontal distance = horizontal velocity(v) x time(t)

6 = 0.6395v

V = 9.4ms^-2 (B is the correct option)

Ans: B

Reference PYQ - Oct/Nov 2016 Paper 11 Q6