Collision between Oldsmobile and Truck: Velocity and Heading Calculation

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The problem involves a collision between a 2082 kg Oldsmobile and a 4070 kg truck, focusing on the calculation of the velocity and heading of the wreckage post-impact. The context includes the application of conservation of momentum principles in a two-dimensional scenario.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss applying conservation of momentum in both x and y directions, with some questioning the angles used in the calculations. There is an attempt to clarify the correct angles for the collision scenario.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the angles involved in the momentum calculations. Some guidance has been offered regarding the application of momentum principles, but no consensus has been reached on the correct setup.

Contextual Notes

There is an indication of urgency due to a homework deadline, and participants are navigating through potential misunderstandings regarding the angles and their implications in the calculations.

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Homework due tomorrow!

A 2082 kg Oldsmobile traveling east on Saginaw Street at 14.3 m/s is unable to stop on the ice covered intersection for a red light at Abbott Road. The car collides with a 4070 kg truck hauling animal feed north on Abbott at 10.8 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg)

I know I've already posted and I'm sorry, but I need someone to help me not so vaguely. I know about the conservation of momentum! Something just isn't adding up.

(2082*14.3)^2 + (4070*10.8)^ = 8.63

angle = 55.9

HELP ME
 
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Apply the principle of conservation of momentum twice, once on the x-resolute and once on the y-resolute.

[tex]m_1v_1\cos\theta+m_2v_2\cos\dot{\theta}=(m_1+m_2)v\cos\ddot{\theta}[/tex]

[tex]m_1v_1\sin\theta+m_2v_2\sin\dot{\theta}=(m_1+m_2)v\sin\ddot{\theta}[/tex]

Where [itex]\theta=45[/itex] and [itex]\dot{\theta}=0[/itex].
 
daster said:
Apply the principle of conservation of momentum twice, once on the x-resolute and once on the y-resolute.

[tex]m_1v_1\cos\theta+m_2v_2\cos\dot{\theta}=(m_1+m_2)v\cos\ddot{\theta}[/tex]

[tex]m_1v_1\sin\theta+m_2v_2\sin\dot{\theta}=(m_1+m_2)v\sin\ddot{\theta}[/tex]

Where [itex]\theta=45[/itex] and [itex]\dot{\theta}=0[/itex].

Shouldn't the angles be 0 and 90 before collision?
 
Yes, sorry. 0 and 90, not 45. :smile:
 

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