there are two cars one traveling north and one traveling east. they both collide.?

Question

they both collide. the white car that is travelling east at (v*A=16.3m/s , m*A=1160kg)

and the black car that is traveling north at (v*B=20.7m/s , m*B=1229kg)

the cars stick together and moce off in another direction. what is the magnitude of the cars' final momenta???
1. 5.96x10^4 kg x m/s
2. 1.19x10^4 kg x m/s
3. 3.13x10^4 kg x m/s
4. 3.17x10^4 kg x m/s


And at what angle north of east is the cars' direction of motion after the collision?? Use Inverse of tan (tan^-1)
1. 36.5 Degrees
2. 38.2 Degrees
3. 39.8 Degrees
4. 53.4 Degrees

Answer 1

Determine the x and y components of the momentum before and after collision. Before collision, they are

px = mA * vA

py = mB * vB

After collision they are

px' = (mA + mB)*vx

py' = (mA + mB)*vy

the components must be equal (px = px'. py = py') so

mA * Va = (mA + mB)*vx

mB * vB = (mA + mB)*vy

Solve for vx and vy:

vx = mA*vA / (mA + mB)

vy = mB*vB / (mA + mB)

The velocity in the direction of motion is v = √[x^2 + y^2];
the momentum is v*(mA + mB).

The angle is arctan[vy/vx]

Plug in the numbers, you get answer (4) in both cases.

NOTE: Sorry I misread you question, and I thought you wanted the velocities. You can get the momentum more easily by computing p = √[px^2 + py^2], and the angle arctan[py/px].

Answer 2

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Answer 3

Vn = (1,160)(16.3)/2389 = 7.9146 m/s
Ve = 1229*20.7/2389 = 10.6489 m/s
θ = tan^-1(7.9146/10.6489) = 36.62°
V = 7.9146/sin36.62° = 13.268 m/s
P = 31,697 kg-m/s = 3.1697*10^4 kg-m/s