9702/Oct Nov/11/2016/Q11

A car has mass m. A person needs to push the car with force F in order to give the car
acceleration a. The person needs to push the car with force 2F in order to give the car
acceleration 3a.

Which expression gives the constant resistive force opposing the motion of the car?

A ma
B 2ma
C 3ma
D 4ma

Solution:

Answer: A

Resultant force = applied force – resistive force

Ma = F- R

R = F- ma —-i

3ma = 2F – R

R = 2F – 3ma —–ii

Substitute for R in eq i

2F – 3ma = F – ma

F = 2ma

Therefore, R = 2ma – ma = ma

Resistive force = ma (A is the correct option)

Reference: PYQ - Oct/Nov 2016 Paper 11 Q11

9702/Oct Nov/13/2013/Q16

The graph shows how the total resistive force acting on a train varies with its speed.

Part of this force is due to wheel friction, which is constant. The rest is due to wind resistance.

What is the ratio wind resistance/wheel friction at a speed of 200 km h^–1?

A 4
B 5
C 8
D 10

Solution:

Answer: A

The wheel friction is constant and the wind resistance increases with speed.

When the speed is zero, the wind resistance is also zero. So, wheel friction = 8 kN.

At a speed of 200 km h^–1, the total resistive force is 40 kN.

Wheel friction = 8 kN

Wind resistance = 40 – 8 = 32 kN

Ratio = wind resistance / wheel friction = 32 / 8 = 4

Reference: PYQ - Oct/Nov 2013 Paper 13 Q16

9702/Oct Nov/11/2009/Q9

The diagram shows two spherical masses approaching each other head-on at an equal speed u.
One has mass 2m and the other has mass m.

Which diagram, showing the situation after the collision, shows the result of an elastic collision?

Solution:
Answer: A

For an elastic collision,

Velocity of approach (before collision) = Velocity of separation (after collision)

{The above result can be obtained by considering that for elastic collision, both momentum and kinetic energy is conserved. Momentum, p = mv. Kinetic energy = ½mv². By equating the sum of momentum before collision to that after collision and by equating the sum of KE before collision to that after collision, 2 equations are obtained which can be simplified into the above stated result: Velocity of approach (before collision) = Velocity of separation (after collision). The proof will not be shown here}

{Approach means that the 2 sphere are coming towards each other and separation means that they are moving away from each other}

Before collision, velocity of approach = u + u = 2u

Consider A:

Velocity of separation = (u/3) + (5u/3) = 6u/3 = 2u

Consider B:

Velocity of separation = (u/6) + (2u/3) = 5u/6

Consider C:

Both spheres are moving in the same direction. So, speed of separation is the difference in the 2 speed here/

Velocity of separation = (2u/3) – (u/6) = 3u/6 = 0.5u

Consider D:

Since the spheres stick together, they are not separating. They move together.

Velocity of separation = 0

Only answer A gives velocity of separation = 2u.

Reference: PYQ - Oct/Nov 2009 Paper 11 Q9

9702/Oct Nov/13/2017/Q10

Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together.

What is the speed of the trucks after the collision?

Solution:
Answer: A

In order to solve this question we need to use the principle of conservation of momentum which states:

The momentum in a closed system remains constant before and after a collision or explosion.

I.E.

momentum before=momentum after

And ingeneral momentum is calculated using: P=mv where P is the momentum, m is the mass of the object, and v is the velocity of the object.

Hence the total momentum before the collision is:

P1+P2
=(m x 2v) + (3m x (-v))

NOTE: notice the negative v on the second truck as it is moving in the opposite direction to the first truck
=-mv

After the collision the the trucks stick together so the total mass becomes 4m and the combined trucks move at an unknown speed of v_​after

We will solve the equation for conservation of momentum to determine v_after​:

mv=4mv_after

cancelling out the m's:

v=4v_after

and rearranging to make v_​after the subject of the equation:

v_after​=0.25v

Which is our final answer.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q10

9702/Oct Nov/13/2017/Q9

A snooker ball of mass 200 g hits the cushion of a snooker table at right-angles with a speed of 14.0ms^–1.

The ball rebounds with half of its initial speed. The ball is in contact with the cushion for 0.60s.

What is the average force exerted on the ball by the cushion?

A.            2.3 N

B.            7.0 N

C.            2300 N

D.            7000 N

Solution:

Answer: B.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q9

9702/May Jun/12/2015/Q14

A ladder is positioned on icy (frictionless) ground and is leant against a rough wall. At the instant
of release it begins to slide.
Which diagram correctly shows the directions of the forces P, W and R acting on the ladder as it
slides?

Solution:

Answer: B.

W is the weight and always acts downwards.

R is the normal reaction to the weight and acts upwards (since ground is frictionless).

For this question, you need to imagine this as a real case – how the ladder would move when it slides.

Before the ladder begins to slide, there is a horizontal force acting to the right at the point of contact of the ladder with the rough wall since the ladder itself exerts a force to the left on the wall due to its mass.

(Diagram A corresponds to this [case where no friction acts and the ladder is not sliding] – The angle of the weight with the ladder is not zero, so there is a component of the weight acting against wall. From Newton’s 3^rd law, there is a reaction from the wall – this is P before the ladder begins to slide).

As the ladder begins to slide, the point of contact of the ladder with the rough wall tends to move downwards. Since the wall is rough friction would act upwards, opposing the motion. So, at the point of contact, there is now a force to the right, along with a force acting upwards, the resultant of which is shown by P in diagram B.

The ground is frictionless, so there is no additional force opposing the motion.

Reference: PYQ - May/Jun 2015 Paper 12 Q14

9702/May Jun/12/2015/Q15

A uniform solid block has weight 500 N, width 0.4 m and height 0.6 m. The block rests on the edge
of a step of depth 0.8 m, as shown.

The block is knocked over the edge of the step and rotates through 90° before coming to rest with

the 0.6 m edge horizontal.

What is the change in gravitational potential energy of the block?

A 300 J 

B 400 J 

C 450 J 

D 550 J

Solution:

Answer: C.

Since the block is said to be uniform, the centre of mass can be considered to be at the centre.

At the top of the step,

Height of position of centre of mass above ground = 0.8 + (0.6/2) = 0.8 + 0.3 = 1.1 m

At the bottom of the step,

Height of position of centre of mass above ground = 0.4/2 = 0.2 m

Change in GPE = mgΔh = 500 × (1.1 – 0.2) = 450 J

Reference: PYQ - May/Jun 2015 Paper 12 Q15

9702/Oct Nov/13/2015/Q11

Question 11

A rocket of mass 30 000 kg sits on a launch pad on the Earth’s surface. The rocket motors
provide an upward force of 330 kN on the rocket.

What is the initial acceleration of the rocket?

A     0.12 m s^–2
B     1.1 m s^–2
C     1.2 m s^–2
D     11 m s^–2

Solution:
30 000kg x 9.81 = 294 300N
330 000 - 294 300 = 35700N
ma = 35700N
a = 35700/30000
   = 1.19
   = 1.2 ms^-2

Ans: C

Reference: PYQ - Oct/Nov 2015 Paper 13 Q11