Category: Physic Tutorials

Simplifying physics for better understanding for students to excel in ‘A level’ and UTME

How to solve questions on Measurement and Kinematics for Cambridge A level

The two questions are cambridge may/june 2012 p21

Q1

The volume V of liquid that flows through a pipe in time t is given by the equation

where P is the pressure difference between the ends of the pipe of radius r and length l.
The constant C depends on the frictional effects of the liquid.
Determine the base units of C.

solution

The base unit of pressure is

the base unit for radius = m

the base unit for length = m

volume = m^3

t = s

the first thing to do is to make C the subject of the equation

Substitute all the base units

Q2

A ball is thrown vertically down towards the ground with an initial velocity of 4.23 m s–1. The
ball falls for a time of 1.51 s before hitting the ground. Air resistance is negligible.
(a) (i) Show that the downwards velocity of the ball when it hits the ground is 19.0 m s–1.
(ii) Calculate, to three significant figures, the distance the ball falls to the ground.

(b) The ball makes contact with the ground for 12.5 ms and rebounds with an upwards
velocity of 18.6 m s–1. The mass of the ball is 46.5 g.
(i) Calculate the average force acting on the ball on impact with the ground

Solution

v = u + at

u = 4.23 ms-1

t = 1.51 s

a = 9.81

v = 4.23 + 1.51 *9.81

v = 19.04 ms-1

ii

s = 17.57 m

b

F = 140 N

A ball of mass 400 g is thrown with an initial velocity of 30.0 m s–1 at an angle of 45.0° to the
horizontal, as shown in fig below

Air resistance is negligible. The ball reaches a maximum height H after a time of 2.16 s.
(i) Calculate
1. the initial kinetic energy of the ball,
2. the maximum height H of the ball

cambridge past que may / jun 2014 p22 q4

Solution

Ek = 1/2 m v2

m = 400g = 0.4kg

v = 30ms-1

Ek = 1/ 2 * 0.4 * 30 ^2 = 180 J

To calculate the maximum

v²=u²– 2as

v = 0

u is the vertical component of the velocity = usinθ = 30 sin 45 = 21.21 ms-1

s = 21.21^2 / 2* 9.81 = 22.94m

2.

A ball is thrown from a point P, which is at ground level, as illustrated in figure below

The initial velocity of the ball is 12.4 m s–1 at an angle of 36° to the horizontal.
The ball just passes over a wall of height h. The ball reaches the wall 0.17 s after it has been
thrown.
Assuming air resistance to be negligible, calculate
(i) the horizontal distance of point P from the wall,
(ii) the height h of the wall.

cambridge past que oct / nov 2010 p22 que 2

Solution

the horizontal distance is

v = d/t

v = ucosθ = 12.4 cos 36 = 10.03 ms-1

d = vt = 10.03 * 0.17 = 1.7 m

the heignt h

s = ut – 1/2 a t2

the u here is the vertical component of the initial velocity usinθ = 12.4 sin 36 = 7.29 ms-1

s = 7.29 * 0.17 – 1/2 * 9.81 * 0.17^2 = 1.24 – 0.14 = 1.1 m

3.

An object is projected from a height of 80m above the ground with a velocity of 40ms-1 at an angle of 30o to the horizontal. The time of flight is
A. 16s B. 10s C. 8s D. 4s
[g = 10ms-2]

Solution

T = 2usinθ / g

T = 2 * 40 * sin 30 / 10

T = 40 / 10

T = 4 s

November 23, 2016
Solving questions on deformation of solids, kinematics and dynamics

Q1

A supermarket trolley, total mass 30 kg, is moving at 3.0 m s–1. A retarding force of 60 N is applied
to the trolley for 0.50 s in the opposite direction to the trolley’s initial velocity.
What is the trolley’s new velocity after the application of the force?{cambridge oct/nov 2009 p11}
A 1.0 m s–1 B 1.5 m s–1 C 2.0 m s–1 D 2.8 m s–1

Solution

v +3 = 1

v = -2 ms-1{ -ve shows that speed is in opposite direction}

C is the correct option

Q2

A car is travelling along a straight road at speed v. A hazard suddenly appears in front of
the car. In the time interval between the hazard appearing and the brakes on the car coming
into operation, the car moves forward a distance of 29.3 m. With the brakes applied, the front
wheels of the car leave skid marks on the road that are 12.8 m long, as illustrated in figure below

It is estimated that, during the skid, the magnitude of the deceleration of the car is 0.85 g,
where g is the acceleration of free fall.{Cambridge oct/Nov 2008 p2}
(i) the speed v of the car before the brakes are applied,
(ii) the time interval between the hazard appearing and the brakes being applied.

Solution

Speed of car is V

The skid mark is 12.8 m

So deceleration = 0.85g = 0.85*9.81 ms-2

using

I used negative sign because the car decelerates

If you do your calculation very well

U = 14.61ms-1

To calculate the time

Since the car was travelling at a stead speed

t = 2 s

Q3

A spring is placed on a flat surface and different weights are placed on it, as shown in the figure below

The variation with weight of the compression of the spring is shown below

The elastic limit of the spring has not been exceeded.{Cambridge may/june 2008 p2}

i Determine the spring constant k of the spring.

ii Deduce that the strain energy stored in the spring is 0.49 J for a compression of 3.5 cm

Solution

The spring constant can be calculated by finding the inverse of the slope of the graph

K = 32/ 0.04 = 800Nm-1

Strain energy stored

Q4

The Young modulus of steel is determined using a length of steel wire and is found to have the

value E.

Another experiment is carried out using a wire of the same steel, but of twice the length and half

the diameter.

What value is obtained for the Young modulus in the second experiment?{ Cambridge may/june 2008 p1}

Solution

From the explanation of Young modulus that I can. It does not change with change in length, it only depend on the material it made up of and not the length.

so, the Young modulus will remain the same

C is the correct option

Note

I will highly appreciate if questions can be asked to know if you have gotten atleast some understanding of the four topics taught so far.

November 20, 2016

Category: Physic Tutorials

How to solve questions on Measurement and Kinematics for Cambridge A level

Solving questions on deformation of solids, kinematics and dynamics