A glider of mass 1500 kg is launched from rest on a straight and level track using a catapult.

Question 15

A glider of mass 1500 kg is launched from rest on a straight and level track using a catapult. The graph shows the variation with time of the resultant force.

What is the speed of the glider when the resultant force acting on it reaches zero?

A 133 m s^-1                       B 200 m s^-1                       C 250 m s^-1                       D 267 m s^-1

Reference: Past Exam Paper – November 2015 Paper 11 Q11

Solution:

Answer: A.

From Newton’s second law:

F = Δp / t

F × t = Δp

(F × t) is the impulse.

 

The change in momentum is given by the impulse, i.e. average force × time or the area under the graph.

Area under graph = Δp = F × t = ½ × 40 × 10 = 200 kNs

Δp = 200 000 Ns

Δp = m(Δv)

Dividing by the mass then gives the change in speed.

Initially, the glider is at rest. i.e. initial speed = 0

Δp = m(v–0) = mv       where v is the final speed

mv = 200 000 Ns

Speed v = 200 000 / 1500 = 133 m s^-1

A sky-diver falls vertically from a helicopter and reaches constant (terminal) velocity.

Question 22

A sky-diver falls vertically from a helicopter and reaches constant (terminal) velocity. The graph shows the variation with time t of the speed v of the sky-diver.

Which graph shows the variation with time t of the distance d fallen by the sky-diver?

Reference: Past Exam Paper – June 2018 Paper 12 Q7

Solution:

Answer: C.

The speed of the sky-diver increases at a decreasing rate until it becomes constant (where terminal velocity is reached).

From a distance-time graph, the speed may be obtained by the gradient of the graph.

So, in the distance-time graph, the gradient should increase at a decreasing rate (a curve) and then becomes constant and non-zero (a straight line, but not horizontal). [B and D are incorrect]

From the velocity-time graph of the sky-diver, initially the speed is zero. So, the gradient of the distance-time graph should be zero initially (that is, the graph should be horizontal initally). [C is correct]

Choice A cannot be correct because the non-zero gradient at time t = 0 shows a non-zero velocity. Only graph C has a zero velocity at time t = 0.

A pipe of length 100 cm is open at both ends. A loudspeaker situated at one end of the pipe can emit sound of different wavelengths.

Question 21

A pipe of length 100 cm is open at both ends. A loudspeaker situated at one end of the pipe can emit sound of different wavelengths.

At which wavelength can a stationary wave be produced in the pipe?

A 50 cm                      B 75 cm                      C 150 cm                    D 300 cm

Reference: Past Exam Paper – June 2017 Paper 13 Q26

Solution:

Answer: A. 

The pipe is opened at both ends. So, antinodes would be located at the ends of the pipe.

The fundamental mode consist of an antinode at both ends (antinode-node-antinode (ANA)). In this case (if you draw the waves), half a wavelength occupies the length of the pipe.

λ / 2 = 100 cm

Wavelength λ = 2 × 100 = 200 cm

This is not an available choice.

The next mode (1^st overtone) would consist of antinodes at the ends and an antinode inside the tube (ANANA). In this case, one complete wave occupies the length of the tube.

λ = 100 cm

This is also not an available choice.

The next mode (2^nd overtone) would consist of antinodes at the ends and 2 antinodes inside the tube (ANANANA). In this case, one and a half wavelength occupies the length of the tube.

3 λ / 2 = 100

λ = 200 / 3 = 66.7 cm

This is also not an available choice.

The next mode (3^rd overtone) would consist of antinodes at the ends and 3 antinodes inside the tube (ANANANANA). In this case, one and a half wavelength occupies the length of the tube.

2 λ = 100

λ = 100 / 2 = 50 cm

Alternatively, we could consider a general formula for the wavelength in open-open pipes. (From your notes,) this would be

(for the (n-1)^th overtone)

λ = 2L / n

Trying values of n (= 1, 2, 3, 4, 5, …) would give the answer.