Question

A loaded truck with a mass of 3 tonnes traveling north at 60 km/h collides with a 1 tonne mass truck, traveling east at 90 km/h at a junction. Calculate in which direction and how far the car is dragged by the truck, knowing that the coefficient of kinetic friction at the accident site is 0.5.

Answer 1

`63.4` deg

`19.94` m

Step-by-step explanation:

`M` = mass of the heavier truck = 3 tonnes

`V_(i)` = velocity of truck before collision =
`0 \hat{i} + 60 \hat{j}` kmh⁻¹

`m` = mass of lighter truck = 1 tonne

`v_(i)` = velocity of lighter truck before collision =
`90 \hat{i} + 0 \hat{j}` kmh⁻¹

`V_(f)` = velocity of combination after collision

Using conservation of momentum

`MV_(i)+ m v_(i) = (M + m)V_(f)`

`(3)(0\hat{i} + 60\hat{j}) + (1)(90\hat{i} + 0\hat{j})= (3 + 1) V_(f)`

`V_(f) =22.5\hat{i} + 45\hat{j}`

magnitude of the final speed is given as

`|V_(f)| = \sqrt{22.5^(2)+45^(2)} = 50.31` km/h

Direction is given as

`\theta = tan^(-1)\left ( (45)/(22.5) \right ) = 63.4` deg

`V_(f)` = velocity of combination after collision = 50.31 km/h = 13.98 m/s

`V` = velocity of combination after it comes to stop = 0 m/s

`a` = acceleration =
`\mu _(k) g` = -
`(0.5) (9.8)` = - 4.9 m/s²

`d` = stopping distance

using the equation

`V^(2) = V^(2)_(f) + 2 ad\\\\(0)^(2) = (13.98)^(2) + 2 (- 4.9)d\\`

`d = 19.94` m