Chapter 9

Question 1

Momentum

Answer 1

Is the product of an object mass and velocity
Mass x velocity

P = momentum (kgms -1)
M = mass (kg)
V = velocity (ms -1)

Question 2

Force

Answer 2

Any interaction that can change the velocity of an object.

Question 3

Newton’s first law of motion

Answer 3

An object remains at rest or in uniform motion unless acted on by a force.

Question 4

Newton’s second law of motion l

Answer 4

The rate of change of momentum of an object is proportional to the resultant force on it. In other words the resultant force is proportional to the change of momentum per second.

Question 5

Consider an object of constant mass m acted on by a constant force f.

Answer 5

It’s acceleration causes a change of its speed from initial speed u to speed v in time T without change of direction

Initial momentum = mu
Final momentum = Mv

Force is proportional to the change of momentum per second 
Therefore 
F = change of momentum / time taken 
= mv - mu / T 
= m(v-u) / t = ma

Question 6

1. If m is constant

Answer 6

F = m x Δv / Δt = ma

Where acceleration a = Δv / Δt

Question 7

2. If m changes at a constant rate

Answer 7

As a result of mass being transferred at constant velocity then Δ x (mv)
= v × Δm where Δm is the change of mass of the object
F = v x Δm / Δt where Δm / Δt = change of mass per second

Question 8

Impulse

Answer 8

F x Δt = Δm x v

Question 9

Force - time graph

Answer 9

Force = y axis 
Time = x axis 

Suppose and object of constant mass m is acted on by a constant force f which changes its velocity from initial velocity u to velocity v in time T

F = mxv - mxu / t

The area under the line of a force-time graph represents the change of momentum or the impulse of the force.

Question 10

Rebound impacts equation

Look at 9.2

Answer 10

Ball bounces off wall
Assume no energy loss
Usin0 = v sin0
Ucos0 = Vcos0

F = mxv - mxu / t

= m x v x cos0 - m x u x cos0 / t

= -2 x m x u x cos0 / t

Question 11

Rebound impact

Example

Answer 11

Mass of 5.0x10 - 26kg
Speed of 420ms-1

A) change of momentum
-420 x 5.0x10 - 26
= -2.1x10x-23

B) the force of the molecule

Force = m x v / t

T = 0.22 x 10-9 seconds = 0.22 nanoseconds 
Force = -2.1 x 10 -23 Ns / 0.22 x 10 -9

Impact force = change of momentum / time

-2.1x10-23 / 0.22x10-9
= -9.5x10-14N

Question 12

Newton’s third law of motion

Answer 12

When two objects interact they expect equal and opposite forces on each other

Question 13

Principal of conservation of momentum

Answer 13

For a system of interacting objects the total momentum remains constant, provided no external resultant forces act on the system.

Question 14

Total…

Answer 14

Total momentum before collision = total momentum after collision

Question 15

Conservation of momentum equation

Answer 15

M1 x u1 + m2 x U2 = m1 x v1 + m2 x v2

Question 16

Elastic collision

Answer 16

No energy lost in the collision

No lost of kinetic energy

Question 17

Inelastic collision

Answer 17

Colliding objects have less kinetic energy

Question 18

Worked example
Elastic and inelastic
9.2

Answer 18

Before 
Box1
= 8000kg 
= 3.0ms-1
Heading towards 
Box2
= 5000kg 
A) m1xv1 + m2xv2 =m1xv1 + m2xv2 
V2 = m1xv1 + m2xv2 - m1xv1 / m2

(8000 x 3) + 0 - (8000 x 1) / 5000
= 3.2 ms-1

After
Box 1 now speed of 1.0ms-1
Box 2 is now moving same direction

B) 1/2 x m1xv1² + 1/2 x m2xv2²
1/2 x 8000 x 3²
= 36000J

1/2 x m2xv2²
1/2 x 8000 x 1² + 1/2 x 5000 x 3.2²
= 29600J

36000 - 29600 = 6400J

Question 19

Objects that collide

Answer 19

Objects that collide and couple together undergo inelastic collisions, as
some of the initial kinetic energy is transferred to the surroundings.
To work out whether a collision is elastic or inelastic, the kinetic
energy of each object before and after the collision must be calculated.