Forces and Dynamics

1 A constant mass undergoes negative uniform acceleration when the resultant force acting on it

A increases uniformly and is in the opposite direction of the mass.

B is constant and opposite to the direction of motion of the mass.

C is constant and in the direction of motion of the mass.

D is constant and in the negative direction.

E decreases uniformly and is in the opposite direction of the motion of mass.

Answer is D.

For uniform acceleration, the resultant force must be constant. Negative acceleration just represents the acceleration is in the negative direction and does not imply decrease nor movement opposite to the direction of motion of the mass.

2 Which of the following statements is untrue about Newton’s Third Law of Motion?

A It involves forces which are equal in magnitude and opposite in direction.

B It involves forces which act on different bodies.

C The force of attraction between an electron and a proton in a hydrogen atom is an example of the forces referred to in Newton’s law of motion.

D The two forces are equal and opposite so that the bodies are in equilibrium.

Ans: D

The forces acts on different bodies and hence, they will not keep the bodies in equilibrium.

3 An object is dropped from a high height and is acted upon by two forces: a constant gravitational force and air resistance which is directly proportional to its velocity.

Which one of the following statements about the subsequent motion of the particle is true?

A Its acceleration increases from zero to a maximum and then decreases.

B Its acceleration increases from zero to a maximum.

C Its velocity increases from zero to a maximum and then decreases.

D Its velocity increases from zero to a maximum.

Ans: D

The object will experience a decreasing resultant force which drops to zero and remains at zero. Hence, the velocity will increase from zero to terminal velocity, which is also the maximum velocity it will attain. The acceleration will decrease from maximum to zero.

4.) A helicopter of mass 2.0 x103 kg rises vertically with a constant speed of 35 m s-1. Taking the acceleration free fall as 10 m s-2, what is the resultant force that is acting on the helicopter?

A zero

B 3.0 x 104 N downwards

C 4.5 x 104 N upwards

D 7.5 x 104 N upwards

E 10.5 x 104 N upwards

Answer is A.

Since speed is constant, resultant force is zero.

5.) A 20 g object traveling to the right at 6.0 m s-1 collides head on with a 30 g object traveling to the left at 7.0 m s-1. What is the loss in kinetic energy for the collision if it was perfectly inelastic?

A 0.01 J

B 1.01 J

C 0.01 kJ

D 1.01 kJ

Ans: B

Loss in kinetic energy = Total final kinetic energy – total initial kinetic energy

Total initial kinetic energy = 0.5[(0.020)(6.0)2+(0.030)(7.0)2]=1.095 J

By conservation of linear momentum and based on the fact that the collision is perfectly inelastic,

m1 u1+m2 u2= (m1+ m2)v

20(6.0)-30(7.0)=(20+30)v

v=-1.8 m s-1

Total final kinetic energy = 0.5(0.050)(1.8)2= 0.081 J

Loss of kinetic energy = 1.014 J

6.)

A particle X (of mass 8 units) and a particle Y (of mass 2 units) move directly towards each other, collide and then separate. If vx is the change of velocity of X and vY is the change of velocity of Y, what is the magnitude of the ratio vx/vY?

A 1/4 B 1/sq root 2 C 1/2 D 2 E 4

Answer is A.

Applying Conservation of Linear momentum,

mX uX+mY uY =mX vX+mY vY

mX vX- mX uX=-( mY vY-mY uY)

mX vx= mYvY

vx/vY= mY /mX= 2/8 =1/4

7.)

A tennis player serves in a tennis match. The ball has a mass of 0.2 kg and was travelling at 40 ms-1 towards the racket just before being struck horizontally by the racket. Figure below shows the magnitude of the force exerted on the ball by the racket as a function of time.

What is the speed of the ball immediately after being struck?

A 10 ms-1.

B 40 ms-1.

C 50 ms-1.

D 90 ms-1.

Answer: A

Area under graph = change in momentum of ball

200 x 50 x 10-3 = m(v – u)

v= (10/0.2) -40 = 10 ms-1. (take final direction of ball as positive)

8.)

An object immersed in a liquid in a tank, experiences upthrust.

What is the physical reason for this upthrust?

A The density of the body differs from that of the liquid.

B The density of the liquid increases with depth.

C The pressure in the liquid increases with depth.

D the value of g in the liquid increases with depth.

Answer C.

9)

A uniform beam of length L is hinged at X and is supported by a tension rod as shown below.

The force exerted by the wall on the beam is in the direction

A XW B XZ C XY D YX

Answer is C

For equilibrium, the line of action of the three forces must pass through a single point. This is only possible if the force acts in the direction XY. It cannot be YX because there will be no translational equilibrium in the horizontal direction.

10)

An object of mass 3.9 kg is on a horizontal frictionless surface. It is held in equilibrium by three strings with different tensions, as shown below. The diagram is not drawn to scale.

What is the value of the tension, T?

A 3.5 N B 9.8 N C 10.7 N D 26.4 N

Answer is C.

For translational equilibrium in the horizontal direction,

T = 14.0 sin 35° + 15.2 sin 10° = 10.7 N

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