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

[KESTRELx]

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.

 

[Doc Al]

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.

 

[KESTRELx]

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

 

[Doc Al]

Good!

 

[asteriatic]

I would approach this problem by first understanding the concept of conservation of momentum. In this scenario, the total momentum of the system (the two cars) before the collision is equal to the total momentum after the collision. This means that the sum of the individual momentums of the two cars before the collision is equal to the sum of their individual momentums after the collision.

Using this concept, I would set up the following equations:

M(a)V(a) + M(b)V(b) = M(a)V'(a) + M(b)V'(b)

Where M(a) and M(b) are the masses of car A and B respectively, V(a) and V(b) are their velocities before the collision, and V'(a) and V'(b) are their velocities after the collision.

Since we know that car A was traveling at 10 m/s before the collision, we can substitute this value into the equation:

1000(10) + 1400V(b) = 1000V'(a) + 1400V'(b)

Next, we can use the given information that the cars become entangled and move off together with a common velocity v' in the direction indicated. This means that V'(a) and V'(b) are equal and we can substitute V'(a) = V'(b) = v' into the equation:

1000(10) + 1400V(b) = 1000v' + 1400v'

Solving for v', we get v' = 12.5 m/s. This is the common velocity that the two cars move off together with after the collision.

To find the corresponding velocity of car B just before the impact, we can use the fact that the angle between the two cars is 90 degrees. This means that the momentum of car B before the collision is entirely in the horizontal direction. Thus, we can use the equation:

M(b)V(b) = M(b)V'(b)

Substituting in the values we know, we get:

1400V(b) = 1400(12.5)

Solving for V(b), we get V(b) = 12.5 m/s. This is the velocity of car B just before impact, which is equal to the common velocity v' after the collision.

In conclusion, as a scientist, I would use the concept of conservation of momentum to solve this problem and find that