Question

In a demolition derby, a 600kg Audi is traveling at 15m/s 30 degrees west of south. A 700 kg BMW is traveling at 10 m/s 40degrees north of east. They collide. After the collation, the Audi is redirected to 20 degrees north of wests and the BMW is redirected to 50 degrees south east. solve for velocity

Answer 1

Since there is no external force applied, So Using momentum conservation in Horizontal direction

Pix = Pfx

m1*v1ix + m2*v2ix = m1*v1fx + m2*v2fx

v1i = Initial speed of Audi = 15 m/s at 30 deg west of south = 15 m/s at 60 deg South of west (In 3rd quadrant x < 0 & y < 0)

v2i = Initial speed of BMW = 10 m/s at 40 deg North of East

v1f = final speed of Audi = v1f at 20 deg North of west (In 2nd quadrant x < 0 and y > 0)

v2f = final speed of BMW = v2f at 50 deg South of East (In 4th quadrant x > 0 and y < 0)

v1ix = -v1i*cos A = -15*cos 60 deg = -7.5 m/s

v2ix = v2i*cos A = 10*cos 40 deg = 7.66 m/s

v1fx = -v1f*cos 20 deg

v2fx = v2f*cos 50 deg

m1 = mass of Audi = 600 kg, & m2 = mass of BMW = 700 kg

600*(-7.5) + 700*7.66 = 600*(-v1f*cos 20 deg) + 700*(v2f*cos 50 deg)

862 = -563.8155*v1f + 449.9513*v2f

Using momentum conservation in vertical direction

Piy = Pfy

m1*v1iy + m2*v2iy = m1*v1fy + m2*v2fy

v1iy = -v1i*sin A = -15*sin 60 deg = -12.99 m/s

v2iy = v2i*sin A = 10*sin 40 deg = 6.428 m/s

v1fy = v1f*sin 20 deg

v2fy = -v2f*sin 50 deg

Using above values:

600*(-12.99) + 700*6.428 = 600*(v1f*sin 20 deg) + 700*(-v2f*sin 50 deg)

-3294.4 = 205.212*v1f - 536.2311*v2f

Now Solving above two equations:

862 = -563.8155*v1f + 449.9513*v2f

-3294.4 = 205.212*v1f - 536.2311*v2f

Using scientific calculator:

v1f = 4.85756 m/s & v2f = 8.00258 m/s

In three significant figures

v1f = 4.86 m/s = final velocity of Audi

v2f = 8.00 m/s = final velocity of BMW

Let me know if you've any query.