Question

A 4000-kg car bumps into a stationary 6000kg truck. The Velocity of the car before the collision was +4m/s and -1m/s after the collision. What is the velocity of the truck?

Answer 1

-4.33m/s

Step-by-step explanation:

Using the law of conservation of momentum

m1u1+m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are their respective velocities

v is the final velocity

From the question;

m1 = 4000kg

m2 = 6000kg

u1 = 4m/s

u2 = ?

v =-1m/s

Substitute into the formula and get u2

4000(4)+6000u2 = (4000+6000)(-1)

16000 + 6000u2 = -10000

6000u2 = -10000-16000

6000u2 = -26000

6u2 = -26

u2 = -26/6

u2 = -13/3

u2 = -4.33m/s

Hence the velocity of the truck is -4.33m/s