Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The d

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4 years ago

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Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The d

rivers step on the brakes and all automobiles are brought to rest. Red Car: 1000kg, 10m/s Yellow Car: 2,000kg, 5m/s Blue Car: 500kg, 20m/s Light Blue Car: 1,000kg, 20m/s Green Car: 500kg, 10m/s Purple Car: 4,000kg, 5m/s Part A Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. Part B Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them. Part C Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest. Rank from largest to smallest. To rank items as equivalent, overlap them.

Physics

2 answers:

Mumz [18] · Answer 1 · 2022-01-14T23:39:01+03:00

Answer:

Part a)

Purple car = Light Blue car > (Red car = Yellow Car = Blue car) > Green Car

Part b)

Purple car = Light Blue car > (Red car = Yellow Car = Blue car) > Green Car

Part c)

Purple car = Light Blue car > (Red car = Yellow Car = Blue car) > Green Car

Explanation:

Red car

mass = 1000 kg

speed = 10 m/s

Yellow car

mass = 2000 kg

speed = 5 m/s

Blue car

mass = 500 kg

speed = 20 m/s

Light Blue car

mass = 1000 kg

speed = 20 m/s

Green car

mass = 500 kg

speed = 10 m/s

Purple car

mass = 4000 kg

speed = 5 m/s

Part a)

Now we know that momentum of each car is product of mass and velocity

so we will have

Red Car

$P_1 = m v$

$P_1 = (1000)(10)$

$P_1 = 10^4 kg m/s$

Yellow Car

$P_2 = m v$

$P_2 = (2000)(5)$

$P_2 = 10^4 kg m/s$

Blue Car

$P_3 = m v$

$P_3 = (500)(20)$

$P_3 = 10^4 kg m/s$

Light Blue Car

$P_4 = m v$

$P_4 = (1000)(20)$

$P_4 = 2\times 10^4 kg m/s$

Green Car

$P_5 = m v$

$P_5 = (500)(10)$

$P_5 = 5 \times 10^3 kg m/s$

Purple Car

$P_6 = m v$

$P_6 = (4000)(5)$

$P_6 = 2\times 10^4 kg m/s$

So the momentum is given as

Purple car = Light Blue car > (Red car = Yellow Car = Blue car) > Green Car

Part b)

Impulse is given as change in momentum so here we can say that final momentum of all the cars will be zero as they all stops and hence the impulse is same as initial momentum of the car

so the order of impulse from largest to least is given as

Purple car = Light Blue car > (Red car = Yellow Car = Blue car) > Green Car

Part c)

Force is defined as rate of change in momentum

Now let say all cars will stop in same time interval

so we will have

$Force = \frac{impulse}{time}$