Edwin asked in Science & MathematicsPhysics · 5 years ago

# physics help with accident analysis?

Two cars collide at an intersection. Car A, with a mass of 2000 kg , is going from west to east, while car B , of mass 1500 kg , is going from north to south at 17.0 m/s . As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 55.0 ∘ south of east from the point of impact.

Part A

How fast were the enmeshed cars moving just after the collision?

Part B

How fast was car A going just before the collision?

Update:

Two cars collide at an intersection. Car A, with a mass of 1900 kg , is going from west to east, while car B , of mass 1400 kg , is going from north to south at 13.0 m/s . As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65.0 ∘ south of east from the point of impact.

Relevance
• NCS
Lv 7
5 years ago

Look at the post-collision information:

tanΘ = opp / adj = south momentum / east momentum

The south momentum is unchanged from before the collision:

south momentum = 1500kg * 17.0m/s = 25 500 kg·m/s, so

tan55º = 25500kg·m/s / east momentum

east momentum = 17 855 kg·m/s

so the initial east velocity was

Ve = east momentum / 2000kg = 8.93 m/s ◄ A's pre-collision velocity

I'm pretty sure this value is correct, not 6.99 m/s as in the other post.

total momentum

p = √(17855² + 25500²) kg·m/s = 31 130 kg·m/s

and the total mass is 3500 kg, so the enmeshed velocity

v = 31130kg·m/s / 3500kg = 8.89 m/s ◄

so I agree with the other post on this value.

Hope this helps!

• oubaas
Lv 7
5 years ago

Ia = ma*Va = 2,000*Va Kg*m/sec

Iay = 0

Iax = 2000*Va

Ib = mb*Vb = 1500*17 = 25,500 kg*m/sec

Iby = -25,500 kg*m/sec

Ibx = 0

Iax = Iby*arctan-35.0° = 13,983 kg*m/sec

Va = Iax/ma = 13,983 /2000 = 6.99 m/sec... speed of car a prior impact

I = Iby/cos 35° = 31,130 kg*m/sec

V = I/(ma+mb) = 31,130/3,500 = 8.89 m/sec ... speed of enmeshed cars after impact