• Physics 9702 Doubts | Help Page 8

Question 45: [Resistance > Power]

Electric heater consists of three similar heating elements A, B and C, connected as shown.

Each heating element is rated as 1.5 kW, 240 V and assumed to have constant resistance. Circuit is connected to 240 V supply.

(a) Calculate resistance of one heating element.

(b) Switches S₁, S₂ and S₃ may be either open or closed. Complete Fig to show total power dissipation of heater for switches in positions indicated.

S₁                     S₂                     S₃                     Total power / kW

Open               closed              closed             

Closed             closed              open

Closed             closed              closed

Closed             open                open

Closed             open                closed

Reference: Past Exam Paper – June 2008 Paper 2 Q6

Solution 45:

(a)

EITHER Power, P = VI and V = IR              OR P = V² / R

Resistance (= V²/ P = 240² / 1500) = 38.4Ω

(b)

S₁                     S₂                     S₃                     Total power / kW

Open               closed              closed              zero

Closed             closed              open                1.5kW

Closed             closed              closed              3.0kW

Closed             open                open                0.75kW

Closed             open                closed              2.25kW

{A and B are in series and these 2 are in a parallel combination with C and the 240V supply. A, B and C each has a resistance of 38.4Ω. Power = V² / R = 240² / R where R is the total resistance being considered in the circuit. In this case, the power can be related to the p.d. across the heater. [P is proportional to V² – this method is actually quicker than always calculating the total resistance]

Consider the switches. If S₁ is opened, no current would flow in the circuit. So, power = zero. Closing S₂ causes B to be short circuited while opening S₂ allows the resistance of B to be included in the total resistance of the circuit. Opening S₃ causes C to be short circuited.

For case 1, no current flows since S₁ is open. Power = 0. 

For case 2, B and C are short circuited. So, p.d. across A = 240V. Total power = 1.5kW. 

For case 3, only B is short circuited. Since A and C are in parallel, the same p.d. of 240V is across them. Each has power = 1.5kW. Total power = 1.5 + 1.5 = 3.0kW.  

For case 4, only C is short circuited. Since A and B are in series, the p.d. across each of them is 120V each. Since P is proportional to V², when V is halved, P becomes one quarter. So, the power of each heater A and B is (1.5 / 4)kW. Total power = (1.5/4) + (1.5/4) = 0.75kW.

For case 5, none of the heaters is short circuited.  A p.d. of 240V would be across C since it is in parallel with the supply. So, power due to C = 1.5kW. There would be a p.d. of 120V across each heater A and B and as in case 4, total power due to A and B = 0.75kW. Total power = 1.5 + 0.75 = 2.25kW}

Question 46: [Kinematics > Projectile motion]

A marble rolls off a table 4m high at a speed of 10ms^-1.

(a) How far from the edge of the table does it strike the floor?

(b) What is its velocity (magnitude and direction) as it strikes the floor?

Reference: ???

Solution 46:

(a)

Horizontal velocity = 10ms^-1

Vertical height, s = 4m

Take acceleration of free fall, g = 9.81ms^-2

Consider the vertical motion.

Time taken for the ball to reach the ground = t

Initial vertical velocity, u = 0ms^-1

s = ut + ½ at²

4 = 0 + 0.5(9.81)t²

Time t = [4.0 / (0.5 x 9.81)]^0.5 = 0.903s

Assuming air resistance is negligible, horizontal acceleration = 0.

For horizontal motion,

Speed = Distance / Time

Distance from edge of table = Speed x Time = 10 x 0.903 = 9.03m

(b)

When ball strikes the floor,

Horizontal component of velocity = 10ms^-1

Vertical component of velocity (downwards) when ball strikes floor = v

v = u + at = 0 + 9.81(0.903) = 8.858ms^-1

Magnitude of velocity = √(10² + 8.858²) = 13.359 = 13.4ms^-1

Angle of velocity below horizontal = tan^-1(8.858 / 10) = 41.5^o

By convention, angles are read in an anticlockwise direction from the horizontal. The above angle is also the acute angle formed from the direction of the velocity and above the horizontal. But it is not the angle that should be stated (due to the convention).

So, the ball strikes the floor with a velocity of 13.4ms^-1 at an angle of {180 – 41.5 =} 138.5^o from the horizontal.

Question 47: [Kinematics > Projectile Motion]

Ball is thrown against vertical wall. Path of the ball is shown.

Ball is thrown from S with initial velocity of 15.0 m s^–1 at 60.0° to horizontal. Assume air resistance is negligible.

(a) For ball at S, calculate

(i) Horizontal component of velocity:

(ii) Vertical component of velocity:

(b) Horizontal distance from S to wall is 9.95 m. Ball hits wall at P with velocity that is at right angles to wall. Ball rebounds to point F that is 6.15 m from wall. Using answers in (a),

(i) Calculate vertical height gained by ball when it travels from S to P

(ii) Show that time taken for ball to travel from S to P is 1.33 s

(iii) Show that velocity of ball immediately after rebounding from wall is about 4.6 m s^–1

(c) Mass of ball is 60 × 10^–3 kg.

(i) Calculate change in momentum of ball as it rebounds from wall.

(ii) State and explain whether collision is elastic or inelastic.

Reference: Past Exam Paper – November 2011 Paper 21 Q3

Solution 47:

(a)

(i) Horizontal component of velocity = 15cos60^o = 7.5ms^-1 

(ii) Vertical component of velocity = 15sin60^o = 13ms^-1 

(b)

(i)

{Consider vertical motion,}

v² = u² + 2as

{Since the ball hits the wall horizontally, it has no vertical component. So, we consider only the horizontal component of velocity. Also, since air resistance is negligible, the horizontal component of velocity remains unchanged. 

So, (taking the upward direction as +ve) initial velocity, u = (+) 13ms^-1. When ball hits the wall, its velocity is zero. So, final velocity, v = 0. Acceleration due to gravity, g is downwards. So, a = g = (-) 9.81ms^-2.

[For Cambridge A-Level Physics, the magnitude of acceleration due to gravity (on earth) is always taken to be 9.81ms^-2, unless stated otherwise. As for the direction, it depends whether the upward or downward direction is taken to be positive. Acceleration due to gravity is always downwards (towards surface). Here the direction of the acceleration is negative as the positive direction has been defined to be upwards]

The equation becomes: 0 = 13² + 2(-9.81) s}

Vertical height, s = 13² / (2x9.81) = 8.6(1)m

(ii)

{For 1^st case, consider vertical motion: v = u + at. 0 = 13 + (-9.81) t giving t = 13 / 9.81. For the OR case, consider horizontal motion. (There is no acceleration). Speed = Distance / Time, giving time t = distance / speed = 9.95 / 7.5}

Time, t = 13 / 9.81 = 1.326s    OR Time, t = 9.95 / 7.5 = 1.327s

(iii)

{Consider horizontal motion. The time for the ball to return to the ground is the same as in (ii), the time for the upward journey of the ball. [At the instant the ball hits the wall (horizontally), it has no vertical component. In the same way, at the instant the ball rebounds; it still has only a horizontal velocity. So, the (resultant) velocity of the ball at that instant is actually equal to the horizontal component (since the vertical component is zero). There is no need to find the resultant here. We calculate the resultant only when the 2 components of the velocity are not zero.] Velocity = distance / time = 6.15 / 1.33.}

Velocity of ball after rebound = 6.15 / 1.33 = 4.6ms^-1

(c)

(i)

{Take motion towards the wall to be +ve and away from the wall to be –ve. [As explained above, at the instant the ball rebounds, it only has a horizontal component of velocity.] Change in momentum, Δp = mΔv = m(v_f – v_i) where v_f is the final velocity (= - 4.6ms^-1) and v_i is the initial velocity (= (+) 7.5ms^-1).}

Change in momentum of ball = (60x10^-3) [– 4.6 – 7.5] = (-) 0.73Ns

(ii)

The final velocity / kinetic energy is less after the collision OR The relative speed of separation is less than the relative speed of approach. Hence, the collision is inelastic.

Question 48: [Momentum > Rocket]

A rocket of mass M is travelling at a constant speed v away from a stationary observer when it instantaneously ejects one-third of its mass at a speed of 6v towards the observer. What is the new speed of the rocket?

A. 1.5v                        B. 2.0v                        C. 4.5v                        D. 5.0v

Reference: ???

Solution 48:

Define the direction away from observer be the positive direction.

Before ejection of mass,

Sum of momentum before ejection of mass = Mv

After ejection of mass,

Since the one-third mass (M / 3) is moving towards the observer, its speed = – 6v.

New mass of rocket = 2M / 3

Let new speed of rocket = v_f

From the conservation of momentum, sum of momentum before ejection of the mass should be equal to the sum of momentum after ejection of the mass.

Mv = (M/3)(– 6v) + (2M/3)v_f

Mv = – 2Mv + 2Mv_f/3

3Mv = 2Mv_f/3

v_f = (3 x 3)v /2 = 4.5v             [Correct answer is C]

Question 49: [Energy > Water flow]

(a) Distinguish between gravitational potential energy and electric potential energy.

(b) Body of mass m moves vertically through distance h near Earth’s surface. Use defining equation for work done to derive expression for gravitational potential energy change of body.

(c) Water flows down a stream from reservoir and then causes water wheel to rotate, as shown.

As water falls through vertical height of 120 m, gravitational potential energy is converted to different forms of energy, including kinetic energy of water. At the water wheel, kinetic energy of water is only 10% of its gravitational potential energy at reservoir.

(i) Show that speed of water as it reaches wheel is 15 m s^–1.

(ii) Rotating water wheel is used to produce 110 kW of electrical power. Calculate mass of water flowing per second through wheel, assuming that production of electric energy from kinetic energy of the water is 25% efficient.

Reference: Past Exam Paper – November 2011 Paper 21 Q4

Solution 49:

Go to

Distinguish between gravitational potential energy and electric potential energy.