Physics 9702 Doubts | Help Page 91

Physics 9702 Doubts | Help Page 91

Question 467: [Stationary waves]
A suspension bridge is to be built across a valley where it is known that the wind can gust at 5s intervals. It is estimated that the speed of transverse waves along the span of the bridge would be 400 ms^-1. The danger of resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length of
A 2000m B 1000m C 400m D 80m E 40m

Reference: Past Exam Paper – N79 / II / 14 & J85 / I / 12

Solution 467:
Answer: B.
Let L be the length of the suspension bridge and the λ be the wavelength of the transverse waves along the span of the bridge.

The fundamental frequency is the lowest frequency of a waveform. This is similar to a sound wave in a tube closed at both ends. It is also called the first harmonics – that is a node would be present at each end of the tube and there’s no other node in between. The length of the tube would then be equal to the length of λ/2 of the wave formed.

The wind can gust at 5s intervals. So, its period is 5s.
Frequency f of the wave = 1/5 = 0.2Hz
Speed v of wave = fλ = 400ms^-1
Wavelength λ = v / f = 400 / 0.2 = 2000m.

So, the danger of resonant motions in the bridge at its fundamental frequency would be greatest if the span had a length equal to λ/2.

Length L = λ / 2 = 2000 / 2 = 1000m

Question 468: [Kinematics]
A double-ended launching device fires two identical steel balls X and Y at exactly same time. Diagram shows the initial velocities of balls. They are both launched horizontally, but Y has greater speed.

Which statement explains what an observer would see?
A Both X and Y reach the ground simultaneously, because air resistance will cause both to have the same final speed.
B Both X and Y reach the ground simultaneously, because gravitational acceleration is the same for both.
C X reaches the ground before Y, because X lands nearer to the launcher.
D Y reaches the ground before X, because Y has greater initial speed.

Reference: Past Exam Paper – June 2013 Paper 13 Q7

Solution 468:

Answer: B.

Both X and Y reach the ground simultaneously, because the acceleration due to gravity, which is downwards, causes them to have the same vertical component of velocity. [B is correct]

Air resistance always opposes motion and increases with the speed of the object. The speed of Y is initially greater than that of X, so air resistance will not cause both of them to have the same final speed. [A is incorrect]

How close or how far the balls fall from the table depends on their horizontal speed, not the vertical speed. The vertical speed affects how fast the balls travel from their vertical height to the ground. [C and D are incorrect]

Question 469: [Dynamics > Momentum]

2.0 kg mass travelling at 3.0 m s^–1 on frictionless surface collides head-on with a stationary 1.0 kg mass. Masses stick together on impact.

How much kinetic energy is lost on impact?

A zero B 2.0 J C 2.4 J D 3.0 J

Reference: Past Exam Paper – June 2013 Paper 11 Q11

Solution 469:

Answer: D.

From the law of conservation of momentum, the sum of momentum before collision should be equal to the sum of momentum after collision.

Linear momentum = mv

Sum of momentum before collision: 2.0 (3.0) + 1.0 (0) = 6.0 Ns

Let the speed at which the masses move after collision be v.

Sum of momentum after collision = (2.0 + 1.0)v = 3.0v Ns

From the law of conservation of momentum,

3v = 6

Speed v = 6 / 3 = 2.0 ms^-1

Kinetic energy, KE = ½ mv²

Sum of KE before collision = 0.5 (2.0) (3.0²) + 0 = 9.0 J

Sum of KE after collision = 0.5 (3.0) (2.0²) = 6.0 J

KE lost = 9 – 6 = 3 J

Question 470: [Kinematics > Air resistance]

Body falls from rest in a vacuum near Earth’s surface. Variation with time t of its speed v is shown below.

Which graph shows variation with time t of the speed v of same ball falling in air at the same place on Earth?

Reference: Past Exam Paper – November 2002 Paper 1 Q10

Solution 470:

Answer: D.

In vacuum, the only force acting on the body is the gravitational force due to Earth. This causes the ball to fall with a constant acceleration of about 9.81 ms^-2. Thus, the gradient of the v-t graph is constant.

Air resistance always opposes motion. So, the net force on the ball is now less than before, causing the acceleration of the ball to be less than 9.81 ms^-2. That is, the increase in speed is now less than before. This means that the new graph cannot have speeds greater than in the previous case at the same time t. [B and C are incorrect]

But air resistance also increases with an increase in speed. Since the speed is continuously increasing (due to the force of gravity), the air resistance also increases. This causes the net force on the ball to decrease with time. Thus, the acceleration of the ball keeps on decreasing with time. That is, the amount by which the speed of the ball increases keeps on getting smaller with time. So, the new graph cannot have a large portion as a straight line. [A is incorrect]

Question 471: [Dynamics > Momentum]

A moving thorium nucleus ²³⁰₉₀Th spontaneously emits α-particle. Nucleus formed is a radium nucleus ²²⁶₈₈Ra, as shown.

Which statement is correct?

A The kinetic energy of the α-particle equals the kinetic energy of the radium nucleus.

B The momentum of the α-particle equals the momentum of the radium nucleus.

C The total momentum before the emission equals the total momentum after the emission.

D The velocity of the α-particle equals the velocity of the radium nucleus.

Reference: Past Exam Paper – November 2013 Paper 13 Q10

Solution 471:

Answer: C.

For any closed system, the total momentum before emission (or collision) equals the total momentum after the emission (or collision). This is the law of conservation of momentum and it applies for any closed system.

If some energy is lost in the system, the total energy after emission can be different than what it was before emission, but momentum is always conserved. [C is correct]

Choice A is not always true, but may vary. Data has not been provided to make such a conclusion.

If choice B was correct, then the total momentum (which is a vector) after emission would be zero and not equal to the total momentum before emission, which is NOT zero since the thorium was moving.

From the conservation of momentum, (momentum p = mv) an object with a larger mass would usually have a smaller velocity and vice versa.