# In an Automobile Collision a 44 Kilogram Passenger

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Two cars collide with each other. Before the collision, one car (m = 1300 kg) is going north at 30 m/s and the other car (m = 900 kg) is going south at 15 m/s. What is the momentum of the system made up of the two cars after the collision?

#6

+33113

For interest’s sake then, let’s suppose the two cars collide elastically, conserving both energy and momentum.

Momentum before:  1300*30 – 900*15 kg.m/s =

Momentum after: 1300v1 + 900v2

so 1300v1 + 900v2 = 25500   …(1)

Kinetic energy before:  1300*302/2 + 900*152/2 = 686250 J

Kinetic energy after:  1300*v12/2 + 900*v22/2

so 1300*v12/2 + 900*v22/2 = 686250   …(2)

Equations (1) and (2) can now be solved for v1 and v2. (v1 = -75/11 m/s ≈ -6.8 m/s;  v2 = 420/11 m/s ≈ 38.2 m/s).

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#1

+26320

Two cars collide with each other. Before the collision, one car (m = 1300 kg) is going north at 30 m/s and the other car (m = 900 kg) is going south at 15 m/s. What is the momentum of the system made up of the two cars after the collision ?

$$\small{\text{  \boxed{v’_1 = v’_2 = \dfrac {(m_1\cdot v_1 – m_2\cdot v_2 )}{ (m_1+ m_2 ) } } \quad \begin{array}{rcl} m_1 &=& 1300\ \mathrm{kg} \\ v_1 &=& 30\ \frac{\mathrm{m} }{ \mathrm{s} } \end{array}\quad \begin{array}{rcl} m_2 &=& 900\ \mathrm{kg} \\ v_2 &=& 15\ \frac{\mathrm{m} }{\mathrm{s}} \end{array}  }}\\\\\\ \small{\text{  v’_1 = v’_2 = \dfrac {(1300\cdot 30 – 900\cdot 15 )}{ (1300+ 900 ) } \ \dfrac{\mathrm{m} }{\mathrm{s}}  }}\\\\ \small{\text{  v’_1 = v’_2 = \dfrac { 25500 }{ 2200 } \ \dfrac{ \mathrm{m}}{\mathrm{s}} = 11.5909090909\ \dfrac{ \mathrm{m} }{\mathrm{s} }  }}\\\\ \small{\text{  m_1\cdot v’_1 + m_2\cdot v’_2 = 1300 * 11.591 + 900 * 11.591 = 25500\ \frac{\mathrm{kg}\cdot \mathrm{m}}{\mathrm{s}}  }}$$

Read:   Gastar Es Antã³nimo De 1 of 1

#2

+33113

Momentum is a vector quantity.  In the absence of any losses, the momentum of the system after the collision is the same as that before the collision. So, taking north as positive the total momentum of the system is:

1300*30 – 900*15 =  25500 kg.m/s

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After the collision the velocity of the two cars, assuming they are stuck together, is

v = 25500/(1300 + 900)

or v = 11.59 m/s  (positive, so going north)

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#3

+394

A question for Alan or Heureka

Why does this equation set fail? Would this work if this were an elastic collision?

Kinetic energy autoN = (0.5(1300)*(302)) = 585,000J

Kinetic energy autoS = (0.5(900)*(152)) = 101,250J

585,000J + (- 101,250J) = 483,750J

Total mass now as a unit = (1300kg) + (900kg) = 2200kg

Solve for
velocity: Sqr(483750J / 0.5(2200kg)) =
20.97m/s in north direction.

_7UP_

#4

+33113

1. Energy is a scalar, not a vector quantity, so you would add not subtract the energies.

2. If it were an elastic collision they wouldn’t stick together.

3. It would be unusual not to lose a lot of energy (dissipated through frictional heat loss, for example) in a head-on car crash!

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#5

+394

Thank you Alan. I usually understand these intuitively, first, then learn the maths. In this case, it is the other way around. I need to develop a mental picture from the scalar and vector dimensions via the maths.

_7UP_

#6

#7

+394

Thank you, Alan. This is like finally seeing the three dimensional picture in a blended color dot image. The two velocities relate to the solution via the harmonic mean. Not unusual considering they are rates. It sure is easy to see this when they are the right values.

2/((38.2)-1+(6.8)-1) = 11.54

I think the most interesting thing is how the lower energy auto dramatically slows the higher one. I saw similar energy-shedding patterns in high-mass low-velocity vs low-mass high-velocity comparisons.

All of this is very cool. Thank you again.

_7UP_

### In an Automobile Collision a 44 Kilogram Passenger

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