Car impact at 90 degrees to each other (conservation of momentum)

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Homework Help Overview

The problem involves two cars colliding at right angles in an intersection, with given masses and initial velocities. The focus is on determining the common velocity after the collision and the velocity of one car just before impact, utilizing principles of momentum and kinetic energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the conservation of momentum and kinetic energy, with some confusion regarding the nature of the collision (elastic vs. perfectly inelastic). There are attempts to set up equations based on these principles, and questions arise about how to handle the entangled state of the cars.

Discussion Status

Some participants have provided calculations for the common velocity and the pre-impact velocity of one car, while others have pointed out the need to treat momentum as a vector quantity. The discussion reflects a mix of approaches and interpretations without a clear consensus on the best method to proceed.

Contextual Notes

There is mention of missing values for the coefficient of restitution, which affects the classification of the collision. The problem is set in a homework context, implying constraints on the methods that can be used.

KESTRELx
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Homework Statement


Two cars collide at right angle in the intersection of two icy roads. Car A has a mass of 1000 kg and car B 1400 kg. The cars become entangled and moved off together with a common velocity v’ in the direction indicated in figure QB4. If car A was traveling 10 m/s at the instant of impact, find
i) v’ and [13marks]
ii) the corresponding velocity of car B just before impact. [7 marks]

v’ is 25 degrees to the vertical, car A on the vertical axis car b on the horizontal axis


Homework Equations



M(a)V(a) + M(b)v(b) = m(t)v'
Ke=1/2mv^2

The Attempt at a Solution



Ok so i have drawn a few conclutions from this, one that the impact is elastic, as i don't physically have the the values to find the coefficient of restitution and i seem to remember that as they have become entangled nothing has reflected.

This means KE and momentum is conserved, However when canceling the Ke equations down. I'm left with effectivly the same equation just a squared in it.

conservation of momentum
M(a)V(a) + M(b)v(b) = m(t)v'
10000 +1400v(b) = 2400v'

conservation of KE
M(a)V(a)^2 + M(b)v(b)^2 = m(t)v'^2

100000 + 1400v(b)^2 = 2400v'^2

substituting one into the other i get something incredibilly complicated and is unlikely to work.
 
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KESTRELx said:
Ok so i have drawn a few conclutions from this, one that the impact is elastic, as i don't physically have the the values to find the coefficient of restitution and i seem to remember that as they have become entangled nothing has reflected.

This means KE and momentum is conserved, However when canceling the Ke equations down. I'm left with effectivly the same equation just a squared in it.
Since the cars become entangled, the collision is perfectly inelastic. Kinetic energy is not conserved.

conservation of momentum
M(a)V(a) + M(b)v(b) = m(t)v'
10000 +1400v(b) = 2400v'
Don't forget that momentum is a vector quantity. Treat vertical and horizontal components separately.
 
Oh snap.

so:
1000 x 10 = 2400 x cos(25) x v'

10000 / ( 2400 x cos (25) ) = v'

v' =4.6ms^-1

and then reversing this for the horizontal

1400v = 2400 x sin(25) x 4.6
v = (2400 x sin(25) x 4.6) / 1400
v= 3.3ms^-1
 
Good!
 

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