A stone is thrown vertically upwards from a point that is 12 m above the sea. It then falls into the sea below after 3.4 s.

Question 32

A stone is thrown vertically upwards from a point that is 12 m above the sea. It then falls into the sea below after 3.4 s.

Air resistance is negligible.

At which speed was the stone released when it was thrown?

A 3.5 m s^-1                        B 6.6 m s^-1                        C 13 m s^-1                        D 20 m s^-1

Reference: Past Exam Paper – March 2019 Paper 12 Q7

Solution:

Answer: C.

The motion of the stone consists of both an upward and a downward movement. So, we need to be careful with the signs of the vector quantities involved.

Let the upward direction be positive. Any quantity pointing downwards would be negative.

Initial velocity = u       (???)

Take the point at which the stone is released to be the origin.

The sea is at a distance of 12 m below the origin. So, the displacement would be negative.

Displacement s = – 12 m

Acceleration a = – 9.8 m s^-2 (this is also negative as it is downwards)

Time t = 3.4 s

Consider the equation of uniformly accelerated motion:

s = ut + ½ at²

- 12 = 3.4u + (½ × -9.8 × 3.4²)

- 12 = 3.4u = 3.4u – 56.644

3.4u = -12 + 56.644

u = 44.644 / 3.4

Initial speed u = 13.1 m s^-1   

Some gas in a cylinder is supplied with thermal energy q.

Question 34

Some gas in a cylinder is supplied with thermal energy q.

The gas does useful work in expanding at constant pressure p from volume V0 to volume VF, as shown.

Which expression gives the efficiency of this change?

Reference: Past Exam Paper – June 2016 Paper 11 Q16

Solution:

Answer: C.

The gas is supplied with thermal energy q.

(Input) thermal energy = q

Work done by expanding gas = pΔV

Work done by expanding gas = p × (V_final – V_initial) = p (V_F – V_O)

This is the output energy.

Output energy = p (V_F – V_O)

Efficiency = output energy / input energy

Efficiency = p (V_F – V_O) / q

A circuit contains a cell, two resistors of resistances R1 and R2 and a variable resistor X.

Question 33

A circuit contains a cell, two resistors of resistances R1 and R2 and a variable resistor X. The cell has negligible internal resistance.

V1 is the potential difference across the resistor of resistance R1.

I2 is the current through the resistor of resistance R2.

The resistance of X is reduced.

What is the effect on V1 and I2?

V1                                           I2

A          decreases                   decreases

B          decreases                   increases

C         increases                     decreases

D         increases                     increases

Reference: Past Exam Paper – June 2016 Paper 13 Q35

Solution:

Answer: C.

The variable resistor R is in parallel with resistor R₂, and this combination is in series with the resistor R₁.

When the resistance of X is reduced, the equivalent resistance of the parallel combination decreases. This causes the p.d. across the parallel combination to decrease.

The sum of p.d. V₁ and the p.d. across the parallel combination is equal to the e.m.f. of the cell. V = IR. Since the resistance of the combination is reduced, the p.d. across it is also reduced. So, V₁ increases. [A and B incorrect]

As the current reaches the junction of the parallel combination, it splits.

From Ohm’s law: I = V / R

The greater the resistance, the small the current. Since the resistance of X has decreased, more current flow through it. This means that the current I₂ decreases. [C correct]