# school exam (lower 6)

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## school exam (lower 6)

1(a) Given the expression X=ρAѵ where ρ ,A, ѵ represent density,

cross-sectional area and velocity respectively. What quantity does X

represent?(2m)

Solution:

[ρAѵ]=ML-3xL2xLT-1

=M/T

X=The rate of change of mass

(b) A quantity Z is defined according to the following expression

Z= πpr4/8ПL

Where k is a dimensionless constant and p,r,П,and L represent the

pressure difference between the ends of a tube, radius of tube,

viscosity and the length of tube. What physical quantity does Z

represent?

Solution: (2m)

R.H.S=[ML-1T-2][L4]/[ML-1T-1][L]

=L3T-1

Z=rate of change of volume

2(a) Define the term acceleration.

Acceleration is the rate of change of velocity.

(b)A force F of 600 N is applied to accelerate two bodies that are in contact with each other as shown in the diagram below.

F

→

If the friction between the bodies and the floor is negligible, what force does the 100kg mass exert on the 200kg mass?(3m)

Solution: F−F y on X =mxa

600−F y on X=100(2)

F y on X=400N

3(a )State the conditions for the equilibrium of a particle.(1m)

The net force acting on a particle must be equal to zero.

4 A particle of mass m moves in a horizontal circle of radius r at

constant tangential speed ѵ, as shown in the figure below.

B

A

When the particle moves from point A to point B where AB is the diameter, determine

(a) The work done by centripetal force. (2m)

Centripetal force is perpendicular to the angular displacement. Hence, the work done =o

(b) The change of linear momentum of the particle(3m)

ΔP=P2−P1=mv2−mv1

=m(-v1)− mv1

=-2 mv1

5(a)A wheel has moment of inertia 0.04kg m2 about an axis which passes

through the centre of the wheel and is perpendicular to the plane of

the wheel. It rotates at angular velocity 600 rpm. At t=0, a retarding

force is applied on the wheel, which accelerates at constant rate and

stops at t=10s.Determine

(a) The angular acceleration of the wheel (1m)

W0=600X2π/60=20πrads-1

α=w- w0/t=0-20π/10=-2πrads-2

(b) The number of revolutions turned through by wheel in 10s.(2m)

W2=W02+2αѳ

0=(20π)2+2(-2π)ѳ

Ѳ=400π2/4π=100π

100π/2π=50 rev

(c ) the torque acting on the wheel while it decelerates.(2m)

τ=Iα=0.04(-2π)=-0.25Nm

cross-sectional area and velocity respectively. What quantity does X

represent?(2m)

Solution:

[ρAѵ]=ML-3xL2xLT-1

=M/T

X=The rate of change of mass

(b) A quantity Z is defined according to the following expression

Z= πpr4/8ПL

Where k is a dimensionless constant and p,r,П,and L represent the

pressure difference between the ends of a tube, radius of tube,

viscosity and the length of tube. What physical quantity does Z

represent?

Solution: (2m)

R.H.S=[ML-1T-2][L4]/[ML-1T-1][L]

=L3T-1

Z=rate of change of volume

2(a) Define the term acceleration.

Acceleration is the rate of change of velocity.

(b)A force F of 600 N is applied to accelerate two bodies that are in contact with each other as shown in the diagram below.

F

→

If the friction between the bodies and the floor is negligible, what force does the 100kg mass exert on the 200kg mass?(3m)

Solution: F−F y on X =mxa

600−F y on X=100(2)

F y on X=400N

3(a )State the conditions for the equilibrium of a particle.(1m)

The net force acting on a particle must be equal to zero.

4 A particle of mass m moves in a horizontal circle of radius r at

constant tangential speed ѵ, as shown in the figure below.

B

A

When the particle moves from point A to point B where AB is the diameter, determine

(a) The work done by centripetal force. (2m)

Centripetal force is perpendicular to the angular displacement. Hence, the work done =o

(b) The change of linear momentum of the particle(3m)

ΔP=P2−P1=mv2−mv1

=m(-v1)− mv1

=-2 mv1

5(a)A wheel has moment of inertia 0.04kg m2 about an axis which passes

through the centre of the wheel and is perpendicular to the plane of

the wheel. It rotates at angular velocity 600 rpm. At t=0, a retarding

force is applied on the wheel, which accelerates at constant rate and

stops at t=10s.Determine

(a) The angular acceleration of the wheel (1m)

W0=600X2π/60=20πrads-1

α=w- w0/t=0-20π/10=-2πrads-2

(b) The number of revolutions turned through by wheel in 10s.(2m)

W2=W02+2αѳ

0=(20π)2+2(-2π)ѳ

Ѳ=400π2/4π=100π

100π/2π=50 rev

(c ) the torque acting on the wheel while it decelerates.(2m)

τ=Iα=0.04(-2π)=-0.25Nm

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