• Physics 9702 Doubts | Help Page 101

Question 518: [Kinematics]
Graph shows how the velocity v of an object moving in straight line varies over time t = 0 to t = T.

Which graph represents displacement s of the object in the time t = 0 to t = T?

Reference: Past Exam Paper – June 2012 Paper 12 Q7



Solution 518:

Answer: A.

The object moves in a straight line.

Velocity is a vector, so a positive value would indicate that the object is moving in one direction along the straight line (e.g. forward) and a negative value would indicate motion in the opposite direction (e.g. backward).

From a velocity-time graph, the displacement is obtained by the area under the graph. Displacement is also a vector, so when the velocity is positive (curve is above the t-axis), the displacement increases and when the velocity is negative (curve is below the t-axis), the displacement decreases.

Interpretation of the velocity-time graph of the object:

Velocity v increases until it reaches a maximum value, and then decreases back to zero (while still having positive values). The increase and decrease are not constant (since graphs are not straight line, so acceleration is not constant). The motion till here is away from the point of start. Thus, the maximum displacement is at this point.

Then, the velocity increases again but now the object is moving in the opposite direction since the velocity has negative values. Since it is mentioned that the motion is in a straight line, the object is now moving towards the point of start. The velocity reaches a maximum value (while still being negative) and then decreases again until it becomes zero. The object has reached its initial point of start.

Question 519: [Work, Energy and Power]
Car travelling on a level road at steady 20 m s^–1 against a constant resistive force develops a power of 40 kW.
What is magnitude of the resistive force?
A 200 N                      B 800 N                      C 2000 N                    D 4000 N

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q18



Solution 519:

Answer: C.

Since the car travels at a steady speed, the net force on the car is zero. So, the resistive force on the car is equal to the force provided by the engine of the car.

Power = Work Done / Time = Force x distance / time = Force x speed

Force = 40 000 / 20 = 2 000 N

Question 520: [Units]
Unit of resistivity, expressed in terms of base units, is given by

kg x³ y^–2 z^–3 .

Which base units are x, y and z?

x          y                      z

A ampere         metre               second
B metre           ampere             second
C metre           second             ampere
D second         ampere             metre

Reference: Past Exam Paper – June 2013 Paper 12 Q2



Solution 520:

Answer: B.

Resistance R = ρL / A

So, resistivity ρ = RA / L

But, from Ohm’s law, resistance R = V / I. Also, power P = VI, giving V = P / I

So, resistance may be written as R = P / I².

Power P = Work done / time = (Force x Distance) / time

Power P = (mass x acceleration x Distance) / time.

Units of power P: kg ms^-2 m s^-1 = kg m² s^-3

So, units of resistance, R: kgm²s^-3A^-2
and units of resistivity ρ (= RA / L): kgm²s^-3A^-2 m² m^-1 = kgm³s^-3A^-2









Question 521: [Work, Energy and Power]
Kinetic energy of vehicle of mass 1000 kg is 4.5 × 10⁵ J. It is braked with total constant braking force of 6000 N.
What will be its stopping distance?
A 37 m                        B 75 m                        C 150 m                      D 300 m

Reference: Past Exam Paper – November 2011 Paper 12 Q16



Solution 521:

Answer: B.

Work done to stop vehicle = Kinetic energy = 4.5x10⁵J

Work done = Force x distance

Distance s = Work done / Force = (4.5x10⁵) / 6000 = 75m

Question 522: [Current of Electricity]

(a) For a cell, explain terms

(i) electromotive force (e.m.f.)

(ii) internal resistance

(b) Circuit of Fig shows two batteries A and B and a resistor R connected in series.

Battery A has e.m.f. of 3.0 V and internal resistance of 0.10 Ω. Battery B has e.m.f. of 12 V and internal resistance of 0.20 Ω. Resistor R has resistance of 3.3 Ω.

(i) Apply Kirchhoff’s second law to calculate current in the circuit

(ii) Calculate power transformed by battery B

(iii) Calculate total energy lost per second in resistor R and the internal resistances

(c) Circuit of Fig may be used to store energy in battery A. Suggest how answers in (b) support this statement

Reference: Past Exam Paper – June 2011 Paper 22 Q5



Solution 522:

(a)

(i) The electromotive force (e.m.f) of a cell is the energy converted from chemical to electrical when charge flows through a cell or round a complete circuit.

(ii) The internal resistance of a cell is (the resistance of a cell) causing loss of voltage or energy loss in the cell.           

(b)

(i)

{From Kirchhoff’s second law, it is known that the algebraic sum of the potential differences in a circuit loop must be zero. In other words, the sum of p.d. across the different components that provide some sort of resistance should be equal to the overall e.m.f. in the circuit.

Current flows from the positive terminal of a battery. If we look at the circuit, it can be seen that the current flowing from the 2 batteries oppose each other. However, in a circuit, current can only flow in one direction. Therefore, in this case, we find the overall e.m.f. that provide a current in ONE direction. This is done by subtracting the e.m.f.s of the batteries. Current would flow from the battery having the higher e.m.f.

The difference in e.m.f.s would be equal to the sum of p.d. across the resistances (including internal resistances).}

E_B – E_A = I (R + r_B + r_A)

12 – 3 = I (3.3 + 0.1 + 0.2)

I = 2.5A

(ii) Power = E_BI = 12(2.5) = 30W

(iii)

(Total resistance, R = 3.3+0.2+0.1 = 3.6. Overall emf, V = 12 – 3 = 9)

Either P = I²R = (2.5)²x3.6 = 22.5Js^-1

Or P = V²/R = 9²/3.6 = 22.5Js^-1

Or P = VI = 9 x 2.5 = 22.5Js^-1

(c) The power supplied from cell B is greater than the energy lost per second in the circuit.