Question

Car A (1750 kg) is travelling due south and car B (1450 kg) is travelling due east. They reach the same intersection at the same time and collide. The cars lock together and move off at 35.8 km/h[E31.6 S]. What was the velocity of each car before they collided?

Answer 1

Consider the east-west direction along x-axis and north-south direction along y-axis. In unit vector notation, velocities can be given as

$\underset{V_{A}}{\rightarrow}$ = velocity of car A before collision = 0 i - $V_{A}$ j

$\underset{V_{B}}{\rightarrow}$ = velocity of car B before collision = $V_{B}$ i + 0 j

$\underset{V_{AB}}{\rightarrow}$ = velocity of combination after collision = (35.8 Cos31.6) i - (35.8 Sin31.6) j = 30.5 i - 18.8 j

$M_{A}$ = mass of car A = 1750 kg

$M_{B}$ = mass of car B = 1450 kg

Using conservation of momentum

$M_{A}$  $\underset{V_{A}}{\rightarrow}$ + $M_{B}$  $\underset{V_{B}}{\rightarrow}$ = ($M_{A}$ + $M_{B}$) ( $\underset{V_{AB}}{\rightarrow}$ )

(1750) (0 i - $V_{A}$ j) + (1450) ($V_{B}$ i + 0 j) = (1750 + 1450) (30.5 i - 18.8 j)

(1450) $V_{B}$ i - (1750) $V_{A}$ j = 97600 i - 60160 j

Comparing the coefficient of "i" and "j" both side

(1450) $V_{B}$ = 97600    and - (1750) $V_{A}$ = - 60160

$V_{B}$ = 67.3 km/h        and  $V_{A}$ = 34.4 km/