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Sergio [31]
3 years ago
15

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

ow down to pull over), the 1300 kg car is slowed to 30 mph and the 1000 kg car is sped up to 31.5 mph. Was momentum conserved in the collision? Was energy conserved in the collision? Show all work. If momentum or energy was lost, calculate how much and discuss where it could have gone.?
Physics
1 answer:
guajiro [1.7K]3 years ago
6 0

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a  = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b  = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

P_1 = m_av_a + m_b v_b

P_1 = 1300 \times 35+ 1000 \times 25

P_1 =70500 Kg.m/s

final momentum

P_2 = m_aV_a + m_b V_b

P_2 = 1300 \times 30+ 1000 \times 31.5

P_2 =70500 Kg.m/s

here  initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

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Gas A has molecules with small mass. Gas B has molecules with larger mass. They are at the same temperature.
Julli [10]

Answer:

c)The gases have the same average kinetic energy.

Explanation:

As we know that the kinetic energy of gas is given as

K = \frac{1}{2}mv^2

here we know that

v = \sqrt{\frac{3RT}{M}}

so we have

K = \frac{1}{2}m (\frac{3RT}{M})

now we have

K = \frac{3}{2}n RT

now mean kinetic energy per molecule is given as

K_{avg} = \frac{3}{2}KT

so this is independent of the mass of the gas

so average kinetic energy will remain same for both the gas molecules

6 0
3 years ago
In a collision, a 15 kg object moving with a velocity of 3 m/s transfers some of its momentum to a 5 kg object. What would be th
Misha Larkins [42]

The key to solve this problem is the conservation of momentum. The momentum of an object is defined as the product between the mass and the velocity, and it's usually labelled with the letter p:

p=mv

The total momentum is the sum of the momentums. The initial situation is the following:

m_A=15,\quad v_A=3,\quad m_B=5,\quad v_B=0

(it's not written explicitly, but I assume that the 5-kg object is still at the beginning).

So, at the beginning, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 3+5\cdot 0=45

At the end, we have

m_A=15,\quad v_A=1,\quad m_B=5,\quad v_B=x

(the mass obviously don't change, the new velocity of the 15-kg object is 1, and the velocity of the 5-kg object is unkown)

After the impact, the total momentum is

p=m_Av_A+m_Bv_B=15\cdot 1+5\cdot x=15+5x

Since the momentum is preserved, the initial and final momentum must be the same. Set an equation between the initial and final momentum and solve it for x, and you'll have the final velocity of the 5-kg object.

4 0
3 years ago
1. A 9000 kg van was stopped at a traffic light when it rear-ended with an 850 compact car moving to the east at a velocity of 5
____ [38]

Answer:

1.785 m/s

Explanation:

The momentum can be calculated using the expression below

M1 *V1 + M2 * V2 = (M1+M2) V3

M1= mass of van=9000 kg

M2= mass of car= 850kg

V3= velocity of entangled car

V1= Velocity of the van= 0

V2= velocity of the car= 5 m/ s

Substitute the values

(900×0) + (500×5)=( 900+500)× V3

2500=1400 V3

V3=2500/1400

V3= 1.785 m/s

Hence, velocity of the entangled cars after collision is 1.785 m/s

8 0
2 years ago
A .5kg bird is perched on its nest so that it has 50J of potential energy. how far is it off the of the ground?
pshichka [43]

It is 10.20 m from the ground.

<u>Explanation:</u>

<u>Given:</u>

m = 0.5 kg

PE = 50 J

We know that the Potential energy is calculated by the formula:

P. E = m \times g \times h

where m is the is mass in kg;  g  is acceleration due to gravity which is 9.8 m/s  and  h  is height in meters.

PE is the Potential Energy.

Potential Energy is the amount of energy stored when an object is stationary.

Here, if we substitute the values in the formula, we get

P. E = m \times g \times h

50 = 0.5 × 9.8 × h

50 = 4.9 × h

h = \frac {50} {4.9}

h = 10.20 m

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