• Physics 9702 Doubts | Help Page 156

Question 772: [Dynamics > Collisions]

Diagram shows a particle P, travelling at speed v, about to collide with a stationary particle Q of the same mass. The collision is perfectly elastic.

Which statement describes motion of P and of Q immediately after the collision?

A P rebounds with speed ½ v and Q acquires speed ½ v.

B P rebounds with speed v and Q remains stationary.

C P and Q both travel in the same direction with speed ½ v.

D P comes to a standstill and Q acquires speed v.

Reference: Past Exam Paper – June 2011 Paper 12 Q13

Solution 772:

Answer: D.

For a perfectly elastic collision, both momentum and energy are conserved.

From the law of conservation of momentum,

Sum of momentum before collision = Sum of collision after collision

Before collision,

Sum of momentum, p = mv + 0 = mv

Total kinetic energy = ½ mv²

After collision,

Consider choice A,

Sum of momentum, p = m(0.5v) + m(-0.5v) = 0.       [A is incorrect]

Consider choice B,

Sum of momentum, p = - mv                                      [B is incorrect] 

Consider choice C,

Sum of momentum, p = m(0.5v) + m(0.5v) = mv

Total kinetic energy = ½ m (0.5v)² + ½ m (0.5v)² = 0.25mv²             [C is incorrect]

Consider choice D,

Sum of momentum, p = m(0) + mv = mv

Total kinetic energy = ½ mv²

Both momentum and energy are conserved.

Question 773: [Waves]

Diagram shows a transverse wave on a rope. Wave is travelling from left to right.

 

At the instant shown, points P and Q on the rope have zero displacement and maximum displacement respectively.

Which of the following describes the direction of motion, if any, of points P and Q at this instant?

point P                         point Q

A         downwards                 stationary

B         stationary                    downwards

C         stationary                    upwards

D         upwards                      stationary

Reference: Past Exam Paper – June 2004 Paper 1Q25

Solution 773:

Answer: A.

The direction of motion of the wave is shown to be to the right in the diagram.

The easiest way to answer this question is to draw, on the diagram itself, the new wave a little bit later in time. This is the same wave as shown, but a little displaced to the right.

This shows that P must go down and that Q barely moves – even if some stated that Q would go down as well, they would not have been able to give that as an answer because it is not there.

Otherwise, it is known that a point at the maximum displacement on a wave is momentarily stationary. At the equilibrium position, the point moves with maximum velocity.

Question 774: [Dynamics]

A resultant force causes body to accelerate.

What is equal to the resultant force?

A the acceleration of the body per unit mass

B the change in kinetic energy of the body per unit time

C the change in momentum of the body per unit time

D the change in velocity of the body per unit time

Reference: Past Exam Paper – June 2014 Paper 12 Q8

Solution 774:

Answer: C.

A resultant force causes body to accelerate.

The resultant force is equal to the change in momentum of the body per unit time (rate of change of momentum) – as stated by Newton’s 2^nd law.

Newton’s 2^nd law: F = Δp / t = m(Δv) / t = ma

In symbol, ‘the acceleration of the body per unit mass’ means (= m / a), which is wrong.

The rate of change of energy (change in energy per unit time) gives the power while the change in velocity of the body per unit time gives the acceleration.

Question 775: [Dynamics > Equilibrium]

Diagram shows an experiment to measure force exerted on a ball by a horizontal air flow.

Ball is suspended by a light string and weighs 0.15 N.

Deflection of the string from vertical is 30°.

What is the force on the ball from the air flow?

A 0.075 N                   B 0.087 N                   C 0.26 N                     D 0.30 N

Reference: Past Exam Paper – November 2012 Paper 11 Q15

Solution 775:

Answer: B.

Consider the ball suspended by the string when there is no air flow. The weight acts vertically downwards. From Newton’s 3^rd law, there is a vertical upward force in the string which is equal in magnitude to the weight. We call this force the ‘tension’.

Now, consider the air flow. This causes a force F, horizontally to the left on the ball. The weight W is always downwards and the tension T is always in the string, which is now 30° to the vertical.

At the maximum displacement, the ball is momentarily at rest. At this position, the force acting on the ball are balanced.

W = T cos30°

F = T sin30°

From the 1^st equation, T = W / cos30°

So, Force F = (W / cos30°) sin30° = W (sin30°/ cos30°) = W tan30°

Force F = 0.15 × tan30° = 0.087 N

Question 776: [Current of Electricity]

In circuit below, a voltmeter of resistance R_V and an ammeter of resistance R_A are used to measure the resistance R of the fixed resistor.

Which condition is necessary for an accurate value to be obtained for R?

A R is much smaller than R_V.

B R is much smaller than R_A.

C R is much greater than R_V.

D R is much greater than R_A.

Reference: Past Exam Paper – June 2014 Paper 11 Q33

Solution 776:

Answer: A.

To obtain the resistance of resistor R from the set-up shown, the potential difference and current flowing through the resistor are measured.

Ohm’s law: V = IR

Resistance = V / I

(Voltmeter and ammeter cannot give resistance on their own. Resistance should be calculated.)

Ideally, a voltmeter should have an infinite resistance and is connected in parallel to the component being investigated. This is to avoid a large current to flow through it. Current is highest when the resistance is smallest.

Ideally, an ammeter should have negligible resistance and is connected in series to the component being investigated so that it does not affect the value of the current flowing.

This, R_V should be very large (larger than R) and R_A should be very small (smaller than R). Hence, the necessary condition for an accurate value to be obtained for R is than R is much smaller than R_V.

The voltmeter must have much greater resistance than R, otherwise there will be significant current in voltmeter and the value obtained from V/I will not be an accurate value for R.

However: If the statement "R_V should be very large (larger than R) and R_A should be very small (smaller than R)" is true, then surely both A and D are correct options?

The idea is that If Rv is too small then the current flows through V and the ammeter reading is not only measuring the current through R. Therefore the equation R=V/I cannot do this as although it is the correct V it is not the I through resistor R - it is instead the current through V and R. 

However, if R_A is large this will have no effect as both I and V will change. Whatever the value of R_A, the voltmeter is only across R and will only measure the V across R. And as the ammeter is in series with R, the ammeter will always measure the current through R. Using R = V/I will give the resistance.

If R_A is large both V and I in R=V/I decrease proportionally and R wouldn't change.