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

AI Thread Summary
Two cars collide at a right angle, with Car A (1000 kg) traveling at 10 m/s and Car B (1400 kg) before becoming entangled. The collision is identified as perfectly inelastic, meaning kinetic energy is not conserved, but momentum is. The conservation of momentum equations lead to a common velocity v' of approximately 4.6 m/s after the collision. The velocity of Car B just before impact is calculated to be about 3.3 m/s. The analysis emphasizes treating momentum as a vector and separating vertical and horizontal components.
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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