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/