# Physics 9702 Doubts | Help Page 30

__Question 183: [Power > Efficiency]__A train on mountain railway is carrying 200 people of average mass 70 kg up slope at angle of 30° to horizontal and at speed of 6.0 m s

^{–1}. Train itself has a mass of 80 000 kg. Percentage of power from the engine which is used to raise the passengers and the train is 40 %.

What is power of the engine?

A 1.1 MW B 2.8 MW C 6.9 MW D 14 MW

**Reference:**

*Past Exam Paper – June 2013 Paper 12 Q17*

__Solution 183:__**Answer: C.**

Total mass in the system = Mass of
passengers + Mass of train

Total mass in the system = 200(70) +
80000 = 94000kg

The weight acts vertically
downwards. [Weight = mg] Since the slope is at an angle of 30

^{o}to the horizontal, the component of the weight acting (down) along the slope is
Component of weight along slope = (94000x9.81)
sin30

**Power required [output] = Work done / time = Force x velocity**

To move at a constant
speed of 6.0ms-1, the force provided by the engine should be equal (and opposite)
to that component of the weight along the slope, such that the resultant
acceleration is zero.

Force required by engine =
(94000x9.81) sin30

Power output by engine = Fv =
[(94000x9.81) sin30] x 6.0 = 2766420W

**Efficiency = Power output / Power input**= 40% = 0.4

Power input = 2766420 / 0.40 = 6.9MW

__Question 184: [Power]__Diagram shows wheel of circumference 0.30 m. Rope is fastened at one end to force meter. Rope passes over wheel and supports a freely hanging load of 100 N. Wheel is driven by an electric motor at constant rate of 50 revolutions per second.

When wheel is turning at this rate, force meter reads 20 N.

What is output power of the motor?

A 0.3 kW B 1.2 kW C 1.8 kW D 3.8 kW

**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q19*

__Solution 184:__**Answer: B.**

The force due to the 100N
load causes the right end of the rope to accelerate downwards (this simultaneously causes the left end to move
upwards). The force due to the 20N at the force meter causes the
left end to accelerate downwards (this simultaneously causes the right end to move upwards). Even if both the 100n and the 20N act downwards, they
have opposite effects on the rope and the direction of turning of the wheel.

Since the 100N load is greater than
the 20N force from the force meter, there is a resultant force acting downwards
at the load (that force would cause the wheel to turn
in a clockwise direction if not balanced) in the rope.

To keep the load from
moving, the wheel should also exert a force equal in magnitude (and in opposite
direction) on the rope [from Newton’s 3

^{rd}law since the resultant force of the rope is 100 – 20 = 90N downwards].
Force exerted by the wheel on the
rope = (100 – 20 =) 80 N

The wheel turns in an anticlockwise
direction at a rate of 50 revolutions per second. The distance moved against
the force in one second is 50 × 0.30 = 15m. [Distance moved in 1
second = speed]

Power provided by motor = Fv = 80 N
× 15 m s

^{–1}= 1200 W = 1.2kW

__Question 185: [Power]__Force of 1000 N is needed to lift hook of a crane at steady velocity. Crane is then used to lift a load of mass 1000 kg at velocity of 0.50 m s

^{–1}.

How much of power developed by motor of the crane is used in lifting the hook and the load? Assume that acceleration of free fall g is equal to 10 m s

^{–2}.

A 5.0 kW B 5.5 kW C 20 kW D 22 kW

**Reference:**

*Past Exam Paper – June 2010 Paper 11 Q15 & Paper 12 Q16 & Paper 13 Q18*

__Solution 185:__**Answer: B.**

For steady velocity, the resultant
force in the system should be zero (resultant
acceleration is zero). So, the force needed to cause the motion of
constant velocity should be to equal to the total force in the system.

Total force in the system
= force of 1000N needed to lift hook+ weight of the load

Weight of the load = 1000 x 10 =
10000N

Total force needed = 10000 + 1000 =
11000N

Power = Work done / time =
Force x distance / time = Force x velocity

Power = (10000 + 1000) x 0.50 =
5.5kW

__Question 186: [Matter > Young modulus]__Composite rod is made by attaching glass-reinforced plastic rod and nylon rod end to end, as shown.

Rods have same cross-sectional area and each rod is 1.00 m in length. Young modulus E

_{p}of the plastic is 40 GPa and Young modulus E

_{n}of the nylon is 2.0 GPa.

Composite rod will break when its total extension reaches 3.0 mm.

What is the greatest tensile stress that can be applied to composite rod before it breaks?

A 7.1 × 10

^{–14}Pa

B 7.1 × 10

^{–2}Pa

C 5.7 × 10

^{6}Pa

D 5.7 × 10

^{9}Pa

**Reference:**

*Past Exam Paper – June 2014 Paper 12 Q21*

__Solution 186:__**Answer: C.**

__total__extension reaches 3.0 mm.

For a material,

**Young modulus, E = stress, S / strain**

Stress = Force / Area

Strain = extension, e / original
length, L

E = S / (e/L) = SL / e

**Extension, e = SL / E**

The same tensile stress (the same force and since the cross-sectional area is the
same for both, the same tensile stress is therefore applied) is applied
to different materials. The different materials will extend by different amounts
when the same stress is applied to each.

**The total extension is 3.0mm (= 3.0x10**.^{-3}m)
For the plastic rod, extension e

_{p}= S (1) / (40x10^{9}) = S / (40x10^{9})
For the nylon rod, extension e

_{n}= S (1) / (2x10^{9}) = S / (2x10^{9})
For the greatest tensile stress that
can be applied to composite rod before it breaks,

e

_{p}+ e_{n}= 3.0x10^{-3}
S/(40x10

^{9}) + S/(2x10^{9}) = 3.0x10^{-3}
Tensile stress, S = (3.0x10

^{-3}) / [1/(40x10^{9}) + 1/(2x10^{9})] = 5.71x10^{6}Pa

__Question 187: [Work, Energy, Power]__A ball released from a height h

_{0}above a horizontal surface rebounds to a height h

_{1}after one bounce. The graph that relates h

_{0}to h

_{1}is shown below.

If the ball (of mass m) was dropped from an initial height h and made three bounces, the kinetic energy of the ball immediately after the third impact with the surface was

A (0.8)

^{3}mgh

B (0.8)

^{2}mgh

C 0.8 mg (h/3)

D [1 – (3x0.2)] mgh

E [1 – (0.8)

^{3}] mgh

**Reference:**

*Past Exam Paper – N81 / II / 2*

__Solution 187:__**Answer: A.**

Initially, potential energy of ball = mgh

h

_{0}is the height before release and h

_{1}is the height after one bounce.

Since the graph is a straight line, the ratio of h

_{1}/ h

_{0}(gradient) is constant.

Gradient = Î”h

_{1}/ Î”h

_{0}= (80 – 0) / (100 – 0) = 0.8

So, after one bounce, the potential energy becomes 0.8 of the initial potential energy.

The kinetic energy of the ball immediately after an impact with the surface is the same as the potential energy at maximum height [all kinetic energy is converted into potential energy].

After 1

^{st}bounce, kinetic energy = 0.8 mgh

After 2

^{nd}bounce, kinetic energy = 0.8 [0.8 mgh] = (0.8)

^{2}mgh

After 3

^{rd}bounce, kinetic energy = (0.8)

^{3}mgh

__Question 188: [Work, Energy, Power]__A mass m moves on a rough plane inclined at an angle Î¸ to the horizontal and, when moving, experiences a constant frictional force F. Mass M is attached to it by means of a light inelastic cord running over a smooth pulley. Mass M is allowed to fall a vertical distance x, causing m to move up the plane as shown in the diagram below.

How much heat is generated by friction in this process?

A Fx

B mgx

C Mgx sinÎ¸

D Mgx sinÎ¸ – Fx

E Mgx sin x + Fx

**Reference:**

*Past Exam Paper – J88 / I / 6*

__Solution 188:__**Answer: A.**

The mass M does not experience any frictional force. Only the mass m, which is on the rough surface, experiences the frictional force F.

When the mass M falls a distance x, mass m also moves a distance x up the plane. Note that this distance is along the surface, in a direction exactly opposite to the frictional force F vector (the distance moved is not at an angle to the force).

The amount of heat generated by the friction is equal to the work done by the frictional force F. {As the mass m moves, it experiences the frictional force. So, here, the frictional force F acts for a distance x – the distance moved by the mass}

Work done = Force x distance moved in direction of the force

Work done = Fx

__Question 189: [Work, Energy, Power]__What is the power required to give a body of mass m a forward acceleration a when it is moving with velocity v up a frictionless track inclined at an angle Î¸ to the horizontal?

A mavg sin Î¸

B mav sin Î¸ + mgv

C mav + mgv sin Î¸

D (mav + mgv) sin Î¸

E (mav + mgv) / sin Î¸

**Reference:**

*Past Exam Paper – N88 / I / 5*

__Solution 189:__**Answer: C.**

Power = Work done / time = Force x velocity

The track is frictionless, so no frictional force acts on the mass.

__If the track was horizontal__, the power required to give the mass m a forward acceleration a when it is moving with velocity v is given by

Power = Force x velocity = (ma) x v = mav

But in this case, the track is inclined at an angle Î¸ to the horizontal. The component of the weight of the mass (against motion) along the track is mg sin Î¸. Therefore, a force is required to keep the mass stationary (from moving down along the inclined track). This force should be equal in magnitude to the component of the weight so that the resultant acceleration (when considering only the component of the weight and this force) is zero.

This force would result in a power of mg sin Î¸ x v = mgv sin Î¸.

**Total power required**= mav + mgv sin Î¸

__Question 190: [Work, Energy, Power]__The mutual potential energy V of two molecules separated by a distance x is shown in the diagram.

Which of the following correctly describes the force between the molecules?

**Reference:**

*Past Exam Paper – J89 / I / 6*

__Solution 190:__**Answer: E.**

There are 2 forces between the two molecules. One force is attractive and predominates when the molecules are far apart and the other force is repulsive and predominates when the molecules are close to each other.

So, there is a particular separation of the molecules when the attractive and repulsive forces balance. That is, there is a separation when the resultant force between the molecules is zero.

The force, F is related to the potential energy, V by

F = - dV / dr

Thus, when the force is zero, dV / dr is zero. That is, the

**of a potential energy, V against separation r graph is zero. This occurs where the graph is either a maximum or a minimum. This point is indicated by r**

__gradient___{2}in the question.

So, for separations less than r

_{2}, the force between the molecules is repulsive and for separations greater than r

_{2}, the force is attractive.

__Question 191: [Work, Energy, Power]__An object of mass m passes a point X with a velocity v and slides up a frictionless incline to stop at point Y which is at a height h above X.

A second object of mass ½ m passes X with a velocity of ½ v. To what height will it rise?

A ¼ h B ½ h C (1/√2) h D h E h√2

**Reference:**

*Past Exam Paper – J87 / I / 5*

__Solution 191:__**Answer: A.**

__For the object of mass m and velocity v__,

At point X, Kinetic energy = ½ mv

^{2}

From conservation of energy,

Potential energy at point Y = Kinetic energy at point X

mgh = ½ mv

^{2}

**Height, h = v**

^{2}/ 2g__For the object of mass m / 2 and velocity v /2__,

At point X, Kinetic energy = ½ (m/2) (v/2)

^{2}= mv

^{2}/ 16

Potential energy at point Y = Kinetic energy at point X

Let the height at point Y = x

(m/2)gx = mv

^{2}/ 16

Height, x = v

^{2}/ 8g = (v

^{2}/ 2g) / 4 = h / 4

__Question 192: [Work, Energy, Power]__The diagram shows two bodies X and Y connected by a light cord passing over alight, free-running pulley. X starts from rest and moves on a smooth plane inclined at 30

^{o}to the horizontal.

What will be the total kinetic energy of the system when X has travelled 2.0m along the plane? (g = 9.8ms

^{-2})

A 20J B 59J C 64J D 132J E 137J

**Reference:**

*Past Exam Paper – J90 / I / 7*

__Solution 192:__**Answer: B.**

There is an acceleration (due to gravity) of 9.8ms

^{-2}downwards on Y since it is hanging freely.

**Weight of Y = mg = 5.0 x 9.8 = 49N**

However, this does not only cause Y to move. X, which is attached to Y will also move. So, the acceleration in the

__whole system__is therefore

**NOT**9.8ms

^{-2}.

__A resultant acceleration (which acts on the total mass in the system) should be calculated__.

It should also be noted that the component of the weight of X will also act on the system. This component acts (downward) along the plane and opposes the direction of motion of X (X should move 2.0 [up] along the slope).

**Component of weight of X along plane = mg sin(30) = 4.0 x 9.8 x sin(30) = 19.6N**

{The weight of Y contributes a force (up) along the slope while the component of weight of X contributes a force (down) along the slope}

Resultant force in system = ma = 49 – 19.6

(4.0 + 5.0) a = 29.4N

Acceleration, a = 29.4 / 9.0 = 3.27ms

^{-2}

This is the resultant acceleration on each X and Y.

Initially both X and Y are at rest.

Consider the equation for uniformly accelerated motion: v

^{2}= u

^{2}+ 2as

Initial speed, u = 0

Final speed = v

Distance travelled, s = 2.0m

Acceleration, a = 3.27ms

^{-2}

v

^{2}= 0

^{2}+ 2(3.27)(2.0)

Final speed, v = √[2(3.27)(2.0)] = 3.62ms

^{-1}

Kinetic energy of X = ½ (4.0) (3.62)

^{2}

Kinetic energy of Y = ½ (5.0) (3.62)

^{2}

Total kinetic energy of system = ½ (4.0) (3.62)

^{2}+ ½ (5.0) (3.62)

^{2}= 58.96 = 59N

For question 187, can't the answer be B? When they say: immediately after third impact with the surface,, don't they mean that we shouldn't include the new third height after that bounce? So, won't there be only 2 height changes: one after the first bounce and one after the second?

ReplyDeleteJust after the 3rd impact, it is the (kinetic) energy that the ball has that will allow it to reach the next maximum height. At this new maximum height, all the energy (previously kinetic) of the ball is now potential energy.

DeleteSo, we can work backwards to find this kinetic energy.

Got it, thank you so much ^__^

ReplyDeleteFor Question

ReplyDelete189 and 191

Why is air resistance ignored?

Frictionless track doesnt mean no air resistance.

No information is given about air, so it is not considered.

Deletethanks to the person who made this blog ..it literally solved all my issues.... where can i find solved papers related to biology and chemistry? i will be very thankful

ReplyDeletei a m very thankful to the person who created this blog it literally solved my problems related to physics...if u can tell where can i find solved papers related to chemistry and biology specially MCQSs i will be very thankful.

ReplyDeletethanks. you may find one for chemistry at

Deletehttp://chemistry-ref.blogspot.com/

but it's not updated.

As for Biology, I do not know.