Momentum

Question 1

momentum is a … quantity

Answer 1

momentum is a vector quantity

Question 2

what is the equation for momentum?

Answer 2

momentum (kgm/s) = mass (kg) x velocity (m/s)

M = m x v

Question 3

rearrange the equation for momentum to make mass the subject

Answer 3

mass = momentum ÷ velocity

m = M ÷ v

Question 4

rearrange the equation for momentum to make velocity the subject

Answer 4

velocity = momentum ÷ mass

v = M ÷ m

Question 5

vector quantities have a … and a …

Answer 5

vector quantities have a magnitude and a direction

Question 6

a car is moving at 15m/s in a northerly direction

it’s mass is 2,000kg

calculate the momentum of the car

Answer 6

M = m x v

= 2000 x 15

= 30, 000 kgm/s north

Question 7

what is the equation for velocity?

Answer 7

velocity (m/s) = distance (m) ÷ time (s)

v = d ÷ t

Question 8

which direction does momentum act in?

Answer 8

in the direction of the velocity

Question 9

explain the statement: the momentum of an object is directly proportional to its mass and velocity

Answer 9

M = m x v

as mass and velocity increase or decrease the momentum will increase or decrease proportionally (by the same factor, or by the same proportion)

Question 10

what does proportionally mean in directly proportional?

Answer 10

by the same factor, or by the same proportion

Question 11

calculate the momentum of an 80kg man running 10m/s

Answer 11

M = m x v

= 80 x 10

= 800 kgm/s

Question 12

calculate the momentum of a 50g ball moving at 80km/h

Answer 12

M = m x v

convert 80km/hour into m/s so x 1000 ÷ (60 x 60) = 80000 ÷ 3600 = 22.2

convert 50g into kg so 50 ÷ 1000 = 0.05

= 0.05 x 2.2

= 1.1 kgm/s

Question 13

a 1300kg car has the momentum of 3900kgm/s acting North

calculate the velocity of the car

Answer 13

M = m x v so v = M ÷ m

= 3900 ÷ 1300

= 3m/s North

Question 14

a hammer is moving at 15m/s and has a momentum of 11.25kgm/s

calculate the mass of the hammer

Answer 14

m = M ÷ v

= 11.25 ÷ 15

= 0.75kg

Question 15

what is the principle of the conservation of momentum?

Answer 15

the principle of the conservation of momentum states that the momentum is always conserved in a collision, provided there are no external forces (e.g. air resisitance, friction) acting on the system

Question 16

give two examples about external forces that could disrupt the principle of the conservation of momentum

Answer 16

air resistance

friction

Question 17

how does this illustrate the conservation of energy?

Answer 17

Before, the velocity is zero. Therefore, the momntum of the dynamte is zero as M = m x v. After, the dynamite explodes in the exact opposite direction (positive and negative). The equal positive and negative momentum cancel each other out having a total momentum of zero. This proves the conservation of momentum as the total momentum before must, and does, equal the total momentum after.

Question 18

A ball which has 5kg mass and is moving east at 10m/s which knocks into another stationairy ball with the mass of 5kg.

Calculate the momentum before and after and calculte the velocity of the balls after

Answer 18

total momentum before = total momentum after

MaVa + MbVb = (Ma +Mb)V

(5 x 10) + (5 x 0) = (5 +5)V

50kgm/s = 50 ÷ (10) = V

5m/s = V

Question 19

A ball which has 5kg mass and is moving east at 10m/s which collides with another ball with the mass of 5kg moving 5m/s west.

Calculate the momentum before and after and then calculat the velocity after

CHECK THE SHEET FOR THIS

Answer 19

total momentum before = total momentum after

MaVa - (backwards direction) MbVb = MaVa + MbVb

(5 x10) - (5x5) = (5 x 2) + (5 x V)

25kgm/s = 10 + (5 x V)

5 x 3 = 15, + 10 = 25

V = 3m/s

Question 20

how do you convert milliseconds into seconds?

Answer 20

÷ 1000

Question 21

how do you convert millimetres into metres?

Answer 21

÷ 1000

Question 22

Use the conservation of momentum to calculate the recoil velocity of the gun:

mass of gun = 1.00kg

mass of bullet = 0.05kg

before: gun and bullet stationairy
after: gun recoils backwards, bullet moves fowards (east) at 300m/s

Answer 22

MgunVgun + MbulletVbullet = MgunVgun + MbulletVbullet

(1.00 x 0) + (0.05 x 0) = (1.00 x ?) + (0.05 x 300)

0 + 0 = (1.00 x ?) + 15

after momentum must equal to 0 as before momentum equals 0

15 ÷ 1.00 = 15

therefore, recoil velocity of the gun = 15 m/s backwards (west)

Question 23

The recoil velocity of a 1.00kg gun is 15 m/s backwards (west)

Explain what would hapen to the recoil velocity if the gun was held by a 99kg person?

Answer 23

the velocity would decrease to x100 smaller as the mass of the gun is now 100kg (1.00kg + 99kg)

the recoil velocity is now 0.15m/s

Question 24

What equation connects force, change in momentum and time taken for change to happen?

Answer 24

force acting (N) = change in momentum (kg m/s) ÷ time taken for change to happen (s)

Question 25

What is Newton’s third law?

Answer 25

if object A exerts a force on object B then object B exerts the excat opposite force on object A

every action has an equal and opposite reaction

Question 26

What does “every action has an equal and opposite reaction” mean?

Answer 26

The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second objec

Question 27

Newtons Second Law (F = ma) to explain how safety features reduce the impact focus in a car crash?

Answer 27

seatbelts, airbags and a cripple zone increase the time taken for a change in momentum so that the impact force decreases

you cannot reduce the change in momentum - it is fixed

the only way to make the force smaller is to make the time bigger as this is an inverse equation

N.B. all always quote and reference this equation

Question 28

What is Newton’s second law?

Answer 28

impact force = (final momentum - initial momentum) ÷ time for change

F = (mv - mu) ÷ t

F = m(v-u) ÷ t

(v-u) ÷ t = acceleration

F = ma

force = mass x acceleration