All external forces on a body cancel out. Which statement must be correct?

Question 13

All external forces on a body cancel out.

Which statement must be correct?

A The body does not move.

B The momentum of the body remains unchanged.

C The speed of the body remains unchanged.

D The total energy (kinetic and potential) of the body remains unchanged.

Reference: Past Exam Paper – June 2013 Paper 12 Q13

Solution:

Answer: B.

All external forces on a body cancel out.

This means that the resultant force on the body is zero.

Resultant force = 0

ma = 0

The resultant force being zero means that the acceleration is also zero.

Acceleration is the rate of change of velocity. So, an acceleration of zero means that the velocity is not changing.

Consider choice A: The body does not move.

A body which is not moving means that its velocity is zero. Acceleration and velocity are two different quantities. No acceleration does not necessarily means that the velocity is zero. It only means that the velocity is not changing. The body may be moving at constant velocity; its acceleration would still be zero.

Consider choice B: The momentum of the body remains unchanged. [Correct]

Momentum p = mv

Since acceleration is zero, the velocity is not changing. So, the momentum is also unchanged.

Consider choice C: The speed of the body remains unchanged.

Speed is only the magnitude while velocity consist of a magnitude (speed) and a direction. When the acceleration is zero, it means that both the speed (magnitude) and the direction are not changing. The speed of a body may be constant but its direction is changing (for example, in circular motion). Then, the acceleration is not zero. So, the speed being unchanged does not necessarily means that the body is not acceleration as we also need to consider the direction.

This is why defining acceleration as ‘rate of change of speed’ is an incorrect definition.

Choice D: The total energy (kinetic and potential) of the body remains unchanged.

Force and energy are two different quantities. So, the resultant force being zero does not equal to the energy being unchanged.

A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end.

Question 22

A spring is attached at one end to a fixed point and hangs vertically with a cube attached to the other end. The cube is initially held so that the spring has zero extension, as shown in Fig. 3.1.

Fig. 3.1                                                            Fig. 3.2

The cube has weight 4.0 N and sides of length 5.1 cm. The cube is released and sinks into water as the spring extends. The cube reaches equilibrium with its base at a depth of 7.0 cm below the water surface, as shown in Fig. 3.2.

The density of the water is 1000 kg m^-3.

(a) Calculate the difference in the pressure exerted by the water on the bottom face and on the top face of the cube. [2]

(b) Use your answer in (a) to show that the upthrust on the cube is 1.3 N. [2]

(c) Calculate the force exerted on the spring by the cube when it is in equilibrium in the water. [1]

(d) The spring obeys Hooke’s law and has a spring constant of 30 N m^-1.

Determine the initial height above the water surface of the base of the cube before it was

released. [3]

(e) The cube in the water is released from the spring.

(i) Determine the initial acceleration of the cube. [2]

(ii) Describe and explain the variation, if any, of the acceleration of the cube as it sinks in the water. [2]

[Total: 12]

Reference: Past Exam Paper – November 2017 Paper 22 Q3

Solution:

(a)

{Pressure in liquid:      P = hρg}

{The top surface is at a depth of (7.0 – 5.1 =) 1.9 cm from the surface.

Calculate the pressure at the top surface and that at the bottom surface and find the difference.}

P = 1000 × 9.81 × 7.0 × 10^-2   or 1000 × 9.81 × 1.9 × 10^-2                     

ΔP = 1000 × 9.81 × (7.0 × 10^-2 – 1.9 × 10^-2) or 686 – 186

{Note that 7.0 cm – 1.9 cm is simply equal to the length of the cube.}

ΔP = 500 Pa                                                                                      

(b)

{Pressure = Force / Area       So, F = PA}

F = pA                         or (Δ)F = Δp × A                                

upthrust = 500 × (5.1×10^-2)² = 1.3 N

or

upthrust = (686 – 186) × (5.1×10^-2)² = 1.3 N

or

upthrust = 1000 × 9.81 × 5.1×10^-2 × (5.1×10^-2)² = 1.3 N

(c)

{For equilibrium, upward forces = downward forces}

{Force exerted by spring + Upthrust = Weight

Force exerted by spring = Weight – Upthrust}

Force exerted by spring= 4.0 – 1.3 = 2.7 N

(d)

{Hooke’s law: F = ke}

extension e = 2.7 / 30

extension = 0.09 (m)   or 9 (cm)

{The extension is 9 cm. But the depth of the base beneath the surface is 7 cm. So,}

height above surface = 9 – 7 = 2 cm

(e)

(i)

{The spring no longer exerts any force on the cube.

Resultant force on cube = Weight – Upthrust = 4.0 – 1.3 = 2.7 N}

{Find the mass of the cube from W = mg}

mass = 4.0 / 9.81

{Resultant force = ma = 2.7 N}

Acceleration = 2.7 / (4.0 / 9.81) = 6.6 m s^-2

(ii)

{Since there is an acceleration, the speed increases. This causes the viscous force to increase. Thus the resultant force on the cube (and hence, the acceleration) decreases until it becomes zero [viscous force + upthrust = weight].

This is the same principle as for terminal velocity}

The viscous force increases (and then becomes constant)

(weight and upthrust constant so,) this causes the acceleration to decrease (to zero)