• Physics 9702 Doubts | Help Page 105

Question 536: [Work, Energy and Power]
Electric motor produces 120 W of useful mechanical output power. Efficiency of the motor is 60 %.
Which row is correct?

electrical power input / W       waste heat power output / W

A                     72                                            48
B                     192                                          72
C                     200                                          72
D                     200                                          80

Reference: Past Exam Paper – June 2011 Paper 12 Q18



Solution 536:

Answer: D.

Useful mechanical output power = 120 W

Efficiency = Useful mechanical output power / Electrical power input = 60%

Input power = Useful power / Efficiency = 120 / 0.6 = 200 W

Waste heat power = Input – Output = 200 – 120 = 80 W

Question 537: [Vectors]

(a)

(i) Distinguish between vector and scalar quantities

(ii) State whether each of the following is vector quantity or scalar quantity.

1. Temperature

2. Acceleration of free fall

3. Electrical resistance

(b) Block of wood of weight 25N held stationary on a slope by means of a string, as shown in Fig. 

Tension in string is T and slope pushes on block with force R that is normal to slope. Tension T in spring

Reference: Past Exam Paper – November 2010 Paper 22 Q1



Solution 537:

(a)

(i) A scalar quantity has only a magnitude while a vector quantity has both a magnitude and a direction.

                                

(ii)

1. Temperature: scalar

2. Acceleration of free fall: vector

3. Electrical resistance: scalar 

(b)

Either

{For the scale diagram, the vector sum of R and the weight should give the magnitude of the tension T, even if the resultant vector obtained would be opposite to T.

To calculate the magnitude of R, we need to consider the triangle formed with R and T are right angles and 25N being the hypotenuse. cos35^o = R / 25 giving R = 25cos35^o = 20.479 = 20.5N

Scale: 1cm represents 5N

So, R would be 4.1cm at 35^o to the vertical and weight of 25N would be 5cm vertically downwards.

The resultant (in green) was obtained, in my case, to be about 2.9cm equivalent to 14.5N. This is within the accepted range.

Note that if the alternative to work by calculation is given, then it’s best to work out the question by calculation.}

A triangle / parallelogram with correct shape

Tension = 14.3N (allow ±0.5N)

Or

{Since R is perpendicular to T, it does not have any component along the slope. So, only a component of the weight balances the tension in the string}

T = 25sin35^o = 14.3N

Question 538: [Kinematics + Momentum]
Ball is released from rest on smooth slope XY.
It moves down slope, along a smooth horizontal surface YZ and rebounds inelastically at Z. Then it moves back to Y and comes to rest momentarily somewhere on XY.

Which velocity-time graph represents motion of the ball?

Reference: Past Exam Paper – November 2011 Paper 11 Q6 & Paper 13 Q7 & November 2014 Paper 13 Q8



Solution 538:

Answer: A.

From X to Y along the slope, the weight of the ball causes it to accelerate uniformly. Thus, the increase in speed in uniform (gradient is constant). From Y to Z, the acceleration of free fall (which is vertically downwards) does not affect the motion of the ball as surface YZ is horizontal. The graph is a horizontal line in this section.

After its inelastic collision (energy is not conserved) with the wall, the kinetic energy (= ½ mv²) of the ball decreases. So, the inelastic collision causes a decrease in speed.  Thus, the ball should take a longer time to return to the slope after the collision (than the time it takes before the collision). These times are represented by the horizontal lines in the graph. [B and D are incorrect]

The gradient of the velocity-time graphs represented the acceleration (and deceleration, where appropriate) of the ball. For graph C, the gradient after the inelastic collision is greater (since it is steeper) than before the collision, indicating that the deceleration is much larger than the acceleration on the friction-free (smooth) slope. This is incorrect since the acceleration due to gravity should be constant. [C is incorrect]

Question 539: [Pressure]

Water in a bath varies in depth from 20.0 cm at shallow end to 30.0 cm at end with the plug.

Density of the water is 1000 kg m^–3.

What is pressure of the water acting on the plug?

A 1960 Pa                   B 2450 Pa                   C 2940 Pa                   D 4900 Pa

Reference: Past Exam Paper – November 2013 Paper 11 & 12 Q22

Solution 539:

Answer: C.

The pressure at a depth h in a liquid of density ρ is given by

Pressure = hρg

Depth of plug = 30.0cm = 0.3m

Pressure acting on plug = hρg = 0.3 x 1000 x 9.81 = 2943Pa