# 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 mv

^{2}
Sum of KE before collision = 0.5 (2.0)
(3.0

^{2}) + 0 = 9.0 J
Sum of KE after collision = 0.5 (3.0)
(2.0

^{2}) = 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

^{230}_{90}Th spontaneously emits α-particle. Nucleus formed is a radium nucleus^{226}_{88}Ra, 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 incorrect]

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.

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