# 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 48B 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 mv

^{2}) 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

This comment has been removed by the author.

ReplyDeleteHi, can qu538 be solved in a different way Considering the distances travelled by the ball after collision, I mean. So on collision it'll lose most of its kinetic energy, hence the distance travelled will be less, as indicated by graph B, since it has lower potential energy now that some kinetic energy is lost.

ReplyDeleteThe horizontal negative line represents the time to move from Z to Y. Since the speed is now lower, this time should be longer on the horizontal positive line (which represents the time traveled to reach Z from Y)

Delete9702 PAPER13 OCTOBER/NOVEMBER 2014

ReplyDeleteQUESTION 16

Go to

Deletehttp://physics-ref.blogspot.com/2015/04/9702-november-2014-paper-13-worked.html

A tennis ball is released from rest at the top of a tall building.

ReplyDeleteWhich graph best represents the variation with time t of the acceleration a of the ball as it falls, assuming that the effects of air resistance are appreciable

Check solution 959 at

Deletehttp://physics-ref.blogspot.com/2015/09/physics-9702-doubts-help-page-198.html