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?

-4.33m/s

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

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.