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Question 262: [Kinematics > Projectile motion]

(a) Stone of mass 56 g is thrown horizontally from the top of cliff with a speed of 18 m s^–1, as illustrated.

Initial height of stone above the level of the sea is 16 m. Air resistance may be neglected.

(i) Calculate change in gravitational potential energy of stone as a result of falling through 16 m.

(ii) Calculate total kinetic energy of the stone as it reaches the sea.

(b) Use answer in (a)(ii) to show that the speed of the stone as it hits the water is approximately 25 m s^–1.

(c) State horizontal velocity of the stone as it hits the water.

(d)

(i) On the grid, draw a vector diagram to represent horizontal velocity and the resultant velocity of the stone as it hits the water.

(ii) Use vector diagram to determine the angle with the horizontal at which the stone hits the water.

Reference: Past Exam Paper – June 2007 Paper 2 Q4



Solution 262:

(a)

(i)

(Change in) gravitational potential energy = mgh = 0.056 (9.8) (16) = 8.78J

(ii) Calculate total kinetic energy of the stone as it reaches the sea:

(Initial) kinetic energy of stone = ½ mv² = 0.5 (0.056) (18)² = 9.07J

Total kinetic energy = 8.78 + 9.07 = 17.9J

(b)

Kinetic energy of stone = ½ mv²

17.9 = 0.5 (0.056) v²               giving speed v = 25(.3) ms^-1   

(c)

Horizontal velocity = 18ms^-1

(d)

(i)

Correct shape of the diagram (2 sides of right-angled triangle with correct orientation)

{In the graph, 5 small squares correspond to 1cm. Since the graph available is more than 9cm, let’s choose a scale of 2cm (10 small squares) to represent 5ms^-1.

First, let’s draw the horizontal velocity.

5ms^-1 is represented by 10squares

18ms^-1 is represented by (18 / 5) x 10 = 36 squares}

The horizontal velocity is the blue line in the diagram.

The resultant velocity vector is obtained by the vector sum of the horizontal velocity and the vertical velocity. The magnitude of the vertical velocity is unknown. So, draw a dotted vertical line (long enough) starting at the head of the horizontal vector and going downwards. This will temporarily represent the horizontal vector (the red dotted line in the diagram).

The resultant vector should start at the tail of the horizontal vector and end at the head of the vertical vector. As shows before, the magnitude of the resultant vector is 25ms^-1.

5ms^-1 is represented by 2cm

25ms^-1 is represented by (25 / 5) x 2 = 10cm

Now, take a ruler and keep the ‘0 cm’ mark at the tail of the horizontal vector. Make sure the length of the rule reaches the dotted vertical line. While always keeping the ‘0 cm’ mark at the tail of the horizontal vector, rotate the ruler until it reaches a point on the dotted vertical line where the distance between that point and the tail of the horizontal vector is 10cm apart. (The point on the dotted line should be at the ’10 cm’ mark on the ruler if you have exactly followed the steps.) So, the resultant vector (shown in green in the diagram) can thus be drawn.}

(ii)

{From the diagram above, I obtained the angle to be 46.1°}

angle with horizontal = 41° → 48°

Question 263: [Dynamics > Momentum]
(a) Define momentum.                        [1]
(b) State Newton’s second law and use it to show that impulse is proportional to the change in momentum.                                 [2]



Solution 263:
(a) Momentum is the product of mass and velocity.

(b) Newton’s second law states that the rate of change of momentum is directly proportional to the force applied.
F = Δp / t
Impulse = F t = Δp









Question 264: [Kinematics > Projectile motion]
Water shots put from a horizontal pipe which is at a height of 52cm from the floor. If the horizontal distance travelled by the water before it hits the floor is 100cm, what is the velocity of the water when it leaves the pipe?

Reference: ???

Solution 264:

The problem is on projectile motion. So, we can consider the vertical and horizontal motion separately. Try to find what is common in the 2 motion. Since water is shot from horizontal pipe, it has no vertical component of velocity.

{As the pipe is horizontal, the moment the water just gets out of the pipe (before gravity starts acting on it - this is an instant, a very very small time), the water will also be moving horizontally. This is a fact. This is just for an instant. For the pipe was at an angle, then the water would have 2 components of velocity.}

Consider the vertical motion.

Distance, s = 0.52m.

Vertical component of initial velocity, u = 0.

Acceleration due to gravity, g = 9.81ms^-2

Time taken for water to hit ground = t (this is the same for both motions)

s = ut + ½ at²

0.52 = 0 + 0.5(9.81)t²

Time, t = √[0.52 / (0.5x9.81)] = 0.3256s

Now, consider the horizontal motion.

Distance travelled, s = 1m

Time taken for travel, t = 0.3256s

Horizontal component of initial velocity = u

Horizontal component of acceleration = 0

s = ut + ½ at²

1 = u(0.3256) + 0

So, horizontal component of initial speed, u = 1 / 0.3256 = 3.07ms^-1

Since the vertical component of initial velocity is zero, the velocity of the water as it leaves the pipe = √(0² + 3.07²) = 3.07ms^-1

Question 265: [Gravitation > Gravitational field strength]
If the Earth has radius R and the acceleration due to gravity at its surface is 9.8 ms^-2, calculate the acceleration due to gravity at a point that is distance r above the surface of a planet with half the radius and the same density as the Earth. (Assume G = 6.7x10^-11 Nm²kg^-2)

Reference: “Calculations for A-level Physics” by T. L. Lowe and J. F. Rounce, 4^th edition, page 71, Exercise 9.2 Q3                                                                                             

Solution 265:
Gravitational field strength, g = GM / r²        [unit: Nkg^-1 or ms^-2]
The numerical value of the gravitational field strength is the same as the acceleration due to gravity. (gravitational field strength is the acceleration due to gravity)

For Earth, g = 9.81ms^-2

Density = Mass, M / Volume
Assuming the planet is spherical,
Volume = 4πR³ / 3

Density ρ = 3M / 4πR³
Mass M = 4πR³ρ / 3

Let mass of Earth = M_E = 4πR³ρ / 3

Let mass of the planet = M_P
The radius of the planet is half that of Earth and its density is the same as that of Earth.
M_P = 4π(R/2)³ρ / 3 = [4πR³ρ / 3] / 8 = M_E / 8
The mass of the planet is 1/8 the mass of Earth.

Gravitational field strength, g = GM / r². G is a constant and the distance r above the surface is also the same in both cases (for Earth and the planet). So, g is directly proportional to the mass of the planet.

For Earth: mass = M_E, g = 9.81ms^-2
For planet: mass = M_E / 8, g = 9.81 / 8 = 1.23ms^-2









Question 266: [Vectors]
Diagram shows two vectors X and Y.

In which vector triangle does vector Z show the magnitude and direction of vector X–Y?

Reference: Past Exam Paper – June 2002 Paper 1 Q2 & November 2007 Paper 1 Q3



Solution 266:
Answer: B.
There are two alternative ways of thinking about vector subtraction. They are; what needs to be added to vector Y to get vector X or, and more common, but less easy to use in practice, vector Z = vector X + (- vector Y).
We will solve the question using 2^nd way.

Z = X – Y = X + (-Y)
The negative of a vector only means that the direction is opposite.
A vector has a ‘head’ and a ‘tail’ and its direction is towards its tail.

Vector addition of 2 vectors is done graphically as follows:
Draw one of the vectors. (In our case, draw vector X first.)

Now, we need to add the second vector (-Y). This is just Y but in the opposite direction.
To perform the addition, we draw the tail of the second vector (that is, the tail of vector (-Y)) at the head of the first vector. However, in the diagrams given, the vector Y is indicated but as said before, (-Y) is just in the opposite direction. Look at the correct choice, B, to better understand.

The resultant vector is draw from the tail of the first vector to the head of the second vector. This is the Z vector in choice B.