In a dodge ball match, two balls collide in mid air. the red ball (0.6 kg) is traveling at 18 m/s, and the blue
1.4 kg) is traveling at 26 m/s. if the red ball bounces backwards at 4 m/s, what is the velocity of the blue
ball after the collision?

← 16.57 ms⁻¹ if red ball initially travels  →

Explanation:

Theory

The law of conservation of linear momentum

The sum of linear momentum of a system under no external force remains constant.

In this scenario although different forces act on two vehicles by each other when we  consider the total system of two cars no external unbalance force act on them. So we can apply this law.

As law says,

The sum of linear momentum before collision = The sum of linear momentum after collision.

Momentum = mass × velocity.

So we get, → 0.6×18 -1.4×26 = -0.6×4 + 1.4×v

where v = velocity of the blue ball,

The velocity of the red ball after collision and the velocity of the blue ball before collision is (-) because they are in ← direction

v =  - 16.57 ms⁻¹ = ← 16.57 ms⁻¹

1 - wave length

2 - amplitude

3 - crest

4 - trough

explanation:

duh.

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s