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lidiya [134]
3 years ago
8

Consider a system to be one train car moving toward another train car at rest. When the trains collide, the two cars stick toget

her. What is the total momentum of the system after the collision?
Before Collison
Car 1 Car 2
m1 =600kg m2 =400kg
v1 =4 m/s v2 =0 m/s
Choices:
800kg × m/s
1,600kg × m/s
2,400kg × m/s
4,000kg × m/s
Someone please help! The lady who tought me did a terrible job of explaining how to do this!
Physics
2 answers:
pickupchik [31]3 years ago
8 0

As per the question, the system consists of two cars.

The the masses of two cars are denoted as-  m_{1} \ and\ m_{2} \ respectively

The two cars undergoes collision with each other. Here collision must be inelastic in nature.

Let the velocities of each car is denoted as-  v_{1} \ and\ v_{2} \ respectively

It is given that masses of two cars and initial velocities-

m_{1} =600 kg          v_{1} =4m/s

m_{2} = 400kg         v_{2} =0 m/s

 

First we have to calculate the total momentum of the system before collision.

The momentum of a body is the product of its mass with velocity.

Momentum of car 1:

                           p_{1} =m_{1} v_{1}

                                  =600kg*4m/s

                                  =2400\ kg\ m\ s^-1

The momentum of car 2:

                             p_{2} =m_{2} v_{2}

                                   =400\ kg*0\ m/s

                                   =\ 0\ kg\ m\ s^-1

Now the total momentum of the system is :

                        p_{1} +p_{2}

                             =[2400+0]\ kg\ m\ s^-1

                             =2400\ kg\ m\ s^-1

Now we are asked to calculate the total momentum of the system after collision.

The basic condition in case of any type of collision is that it is in accordance with law of conservation of linear momentum.

Hence, the momentum will be conserved here.

It means  the total momentum before collision must be equal to the total momentum after collision.

Hence, the correct answer to the question will be 2400 kg m/s.



                 

3241004551 [841]3 years ago
6 0

Answer : 2400 kgm/s.

Explanation :

It is given that,

Mass of train 1, m_1=600\ kg

Mass of train 2, m_2=400\ kg

<em>When the trains collide, the two cars stick together. So, it is a perfectly inelastic collision. The momentum of the system will remain conserved.</em>

Momentum of train car 1, p_1=m_1v_1=2400\ Kgm/s

Momentum of train car 2, p_2=m_2v_2=0

So, total momentum of the system after the collision will be same as before the collision.

p_f=p_1+p_2

p_f=2400\ kgm/s

Hence, after the collision total momentum of the system would be 2400 kgm/s.

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You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Ap
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The answer is 35 minutes

The Newton's law of cooling is:
T(x) = Ta + (To - Ta)e⁻ⁿˣ

T(x) - the temperature of the coffee at time x
Ta - the ambient temperature
To - the initial temperature
n - constant

step 1. Calculate constant k:

We have:
T(x) = 200°F
x = 10 min
Ta = 68°F
To = 210°F
n = ?

T(x) = Ta + (To - Ta)e⁻ⁿˣ
200 = 68 + (210 - 68)e⁻ⁿ*¹⁰
200 = 68 + 142 * e⁻¹⁰ⁿ
200 - 68 = 142 * e⁻¹⁰ⁿ
132 = 142 * e⁻¹⁰ⁿ
e⁻¹⁰ⁿ = 132/142
e⁻¹⁰ⁿ = 0.93

Logarithm both sides with natural logarithm:
ln(e⁻¹⁰ⁿ) = ln(0.93)
-10n * ln(e) = -0.07
-10n * 1 = - 0.07
-10n = -0.07
n = -0.07 / - 10
n = 0.007

Step 2. Calculate time x when T(x) = 180°F:
We have:
T(x) = 180°F
x = ?
Ta = 68°F
To = 210°F
n = 0.007

T(x) = Ta + (To - Ta)e⁻ⁿˣ
180 = 68 + (210 - 68)e⁻⁰.⁰⁰⁷*ˣ
180 - 68 = 142 * e⁻⁰.⁰⁰⁷*ˣ
112 = 142 * e⁻⁰.⁰⁰⁷⁾*ˣ
e⁻⁰.⁰⁰⁷*ˣ = 112/142
e⁻⁰.⁰⁰⁷*ˣ = 0.79

Logarithm both sides with natural logarithm:
ln(e⁻⁰.⁰⁰⁷*ˣ) = ln(0.79)
-0.007x * ln(e) = -0.24
-0.007x * 1 = -0.24
-0.007x = -0.24
x = -0.24 / -0.007
x ≈ 35
5 0
3 years ago
Read 2 more answers
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