*1. A car of mass 1000 kg travelling at 36 km/h is brought to rest over a distance of 20 m. Find:*

(a) the average retardation; (b) the average braking force in newtons.

(a) the average retardation; (b) the average braking force in newtons.

*The retardation is found from the formula,*

v² = u² + 2ax ( where v is the final velocity; u is the initial velocity; a is the acceletation and x is the distance)

v² = u² + 2ax ( where v is the final velocity; u is the initial velocity; a is the acceletation and x is the distance)

**so**

*, v = 0 m/s , u = 36 km/h = 36 × 1000 / 60 × 60 = 10 m/s*

x = 20 m

x = 20 m

**By substitution,**

0² = 10² + 2 × a × 20

0² = 10² + 2 × a × 20

**therefore**

*a = - 10² / 2 × 20 = -2.5 m/s² ( minus sign means retardation) .*

**Knowing**

*m*

**and having found**

*a,*

**we now substitute in**

*to find*

F = ma,

F = ma,

*F.*

**Thus,**

*that is the average braking force*

F= 1000 × 2.5 = 2500 N (

F= 1000 × 2.5 = 2500 N (

*)*

P.s. If you want to know how I typed the multiplication sign (× ) and the superscript ( ² ), that is easy : just hold Alt 0215 to type × and hold Alt 0178 to type ².

*We must first find the average retardation from the formula,*

2. A bullet of mass 20 g, travelling with a velocity of 16 m/s, penetrates a sandbag and is brought to rest in 0.05 s.

Find : (a) the depth of penetration in metres;(b) the average retarding force of the sand in newtons.

2. A bullet of mass 20 g, travelling with a velocity of 16 m/s, penetrates a sandbag and is brought to rest in 0.05 s.

Find : (a) the depth of penetration in metres;(b) the average retarding force of the sand in newtons.

*where*

v = u + at (

v = u + at (

*v=0, u = 16m /s*and

*t =0.05s)*

**Substituting**

*,*

0 = 16 + a × 0.05from which

0 = 16 + a × 0.05

*a = - 16 / 0.05 = - 320 m/s² ( minus sign means retardation) .*

**The depth of penetration may be found using either**

*Suppose we choose the latter, then*

x = ut + 1/2 at²

or

v² = u² + 2ax

x = ut + 1/2 at²

or

v² = u² + 2ax

*whence*

0² = 16² + 2(-320) × x

0² = 16² + 2(-320) × x

*x = -16 × 16 / 2(-320) = 0.4 m*

The average retardation force may now be calculated from

*in which*

F = ma

F = ma

*m = 20g = 0.02 kg ( we must convert to kg)*and

*a = -320 m/s²*

**Thus,**

*F = 0.02 × -320 = - 6.4 N ( minus sign means retardating force).*

3. The valve of a cylinder containing 12 kg of compressed gas is opened and the cylinder empties in 1 min 30 s. If the gas issues from the exit nozzle with an average velocity of 25 m /s,

find the force exerted on the cylinder.

3. The valve of a cylinder containing 12 kg of compressed gas is opened and the cylinder empties in 1 min 30 s. If the gas issues from the exit nozzle with an average velocity of 25 m /s,

find the force exerted on the cylinder.

**The force required to accelerate the gas out of the cylinder is given by**

*,*

F = ma

= mass × change in velocity/ time taken

= change in momentum/ time taken

The velocity of the gas changes from rest to 25 m / s

change in momentum = 12 × 25 kg m/s

time taken = 90 s

average force on gas = 12 × 25/90 = 3.3 N

F = ma

= mass × change in velocity/ time taken

= change in momentum/ time taken

The velocity of the gas changes from rest to 25 m / s

change in momentum = 12 × 25 kg m/s

time taken = 90 s

average force on gas = 12 × 25/90 = 3.3 N

**By Newton's third law, an equal reaction force is exerted on the cylinder**

*;*

**average force on gas = average force on the cylinder**

*= 3.3 N.*

4. A 50g load is placed on a straight air-track sloaping at an angle of 45° to the horizontal. Calculate, in cm/s², the acceleration of the load as it is slides down and also the distance it would move from rest in 0.2 s. (

4. A 50g load is placed on a straight air-track sloaping at an angle of 45° to the horizontal. Calculate, in cm/s², the acceleration of the load as it is slides down and also the distance it would move from rest in 0.2 s. (

**Assume negligible friction and take**

*g = 980 cm/s².)*

**This example illustrates the use of CGS units.**

**The force on the load causing it to accelerate is the resolved part of its weight along the air-track and is equal to**

50 cos 45° gf.

50 cos 45° gf.

**Now**

*, cos 45° = 0.707*and hence

*,*

the accelerating force = 50 × 0.707 × 980 dyn

the accelerating force = 50 × 0.707 × 980 dyn

**The acceleration is found from the equation**

*F = ma*

**thus**

*, a = F / m = 693 cm/s²*

**The distance**

*, x,*

**moved from rest is found from the equation**

*x = ut + 1/2 at²*

where u = 0 cm/s

a= 693 cm/s²

t = 0.2 s

where u = 0 cm/s

a= 693 cm/s²

t = 0.2 s

**Substituting**

*,*

x = 0 × 0.2 + 693 × 0.2² /2

= 13.9 cm.

x = 0 × 0.2 + 693 × 0.2² /2

= 13.9 cm.

**Now, my dear friend, try to solve this problem**

*:*

A man whose mass is 70 kg stands on a spring weighing machine inside an elevator. When the elevator starts to ascend its acceleration is 2.45 m / s². What is reading on the scales? What will the weighing machine read : (a) when the velocity of the elevator is uniform;(b) as it comes to rest with a retardation of 4.9 m/s²? ( Hint: revise post

A man whose mass is 70 kg stands on a spring weighing machine inside an elevator. When the elevator starts to ascend its acceleration is 2.45 m / s². What is reading on the scales? What will the weighing machine read : (a) when the velocity of the elevator is uniform;(b) as it comes to rest with a retardation of 4.9 m/s²? ( Hint: revise post

### Newton's second law of motion