Question

two cars collide at an intersection. Car A, with a mass of 2000 , is going from west to east, while car B, of mass 1500 , is going from north to south at 17.0 . As aresult of this collision, the two cars become enmeshed and move as one afterwards. In your role as an expert witness, you inspect the scene and determine that, afterthe collision, the enmeshed cars moved at an angle of 65.0 south of east from the point of impact.

How fast were the enmeshed cars moving just after the collision??
How fast was carA going just before the collision?

Answer 1

conserving momentum,
2000*v(i) -1500*17 (j) = 3500*(v'(cos65 i - sin65 j))

comparing co-efficients,we get

v' = 6.09 m/s <---speed of the enmeshed cars
so, v= 4.24 m/s <-------speed of car A

Answer 2

This describes a perfectly inelastic collision. Car A was moving only in x and Car B was moving only in y, which is convenient. Car B had y-momentum of(1500 kg)(14 m/s) = 21000 kg-m/s in the south direction, which must therefore also be the y-momentum of the Cars AB enmeshed mass. If the cars' angle ofmotion together was 65 degrees south of east, the cars' momentum must also be in this direction. That momentum has a y-component of 21000 kg-m/s, so, usingtrigonometry, the x-component must be (21000 kg-m/s) / tan(65). Then, knowing the x- and y-components of the momentum, the total momentum can be foundusing the Pythagorean Theorem; it would be the square root of the sum of the squares of the components. Divide that momentum by the combined mass of thecars, 3500 kg, to answer part A. To answer part B, just recognize that the x-momentum of Cars AB is the same as the original momentum of Car A. You alreadyknow that momentum, so simply divide it by the mass of Car A, 2000 kg, to get the car's original speed

Answer 3

East is the +x direction
North is the +y direction

I am more familiar with numeric labeling, so car A is car 1...and car B is car 2.

Initial velocity of each car in terms of speed:
v1i = V1*<1, 0>
v2i = V2*<0, -1>

Total initial momentum:
p = m1*v1i + m2*v2i

The cars have an inelastic collision, where conservation of momentum applies, and end moving at a final velocity vf:
p = (m1 + m2)*vf

Thus:
m1*v1i + m2*v2i = (m1 + m2)*vf

Solve for vf:
vf = (m1*v1i + m2*v2i)/(m1 + m2)

Remember this is a vector equation:
vf = <m1*V1/(m1 + m2), -m2*V2/(m1 + m2)>

In terms of the given angle, it implies that:
tan(theta) = -vfy/vfx

Which means that:
tan(theta) = (m2*V2/(m1 + m2))/(m1*V1/(m1 + m2))

And it simplifies:
tan(theta) = m2*V2/(m1*V1)

So,
V2 = m1*V1*tan(theta)/m2

And we can substitute:
vf = <m1*V1/(m1 + m2), -m1*V1*tan(theta)/(m1 + m2)>

And combine with Pythagorean theorem to find Vf, the post-impact speed:
Vf = sqrt(vfx^2 + vfy^2)
Vf = sqrt((m1*V1)^2 + (m1*V1*tan(theta))^2)/(m1 + m2)

Simplifies:
Vf = m1*V1*sqrt(1 + tan(theta)^2)/(m1+m2)

Summary:
Vf = m1*V1*sqrt(1 + tan(theta)^2)/(m1+m2)
V2 = m1*V1*tan(theta)/m2

Data:
m1:=2000 kg; m2:=1500 kg; V1:=17.0 m/s; theta:=65 deg;

Result:
Vf = 22.99 m/s
V2 = 48.61 m/s, Car B's driver is at fault for driving much faster than any practical road speed limit, that is over 100 mph.