A 20 g ball of clay traveling east at 8.0 m/s collides and sticks together with a 10 g ball of clay traveling north at 7.5 m/s.

What is the speed of the resulting ball of clay?

5.89 m/s

Explanation:

Law of conservation of momentum: It states that in a closed system, if two bodies collides, the total momentum before collision is equal to the total momentum after collision.

From the law of conservation of momentum,

m₁u₁ + m₂u₂ = V (m₁ + m₂) ... Equation 1

Where m₁ = mass of the first ball, m₂ = mass of the second ball, u₁ = initial velocity of the first ball, u₂ = initial velocity of the second ball, V = common velocity.

making V the subject of the equation,

V = (m₁u₁ + m₂u₂) / (m₁ + m₂) ... Equation 2

Note: If both ball stick together after collision, then the collision is inelastic, and as such they move with a common velocity. Both velocities form a right angle triangle and as such the common velocity will have both horizontal and vertical component.

For the horizontal component,

Given: m₁ = 20 g = (20/1000) kg = 0.02 kg, m₂ = 10 g = 10/1000 = 0.01 kg, u₁ = 8.0 m/s, u₂ = 0 m/s (the second ball have a horizontal velocity of 0 m/s)

Substituting into equation 2

V₁ = (0.02*8 + 0.01*0) / (0.02+0.01)

V₁ = (0.16+0) / 0.03

V₁ = 0.16/0.03

V₁ = 5.33 m/s

For the vertical velocity,

u₁ = 0 m/s, u₂ = 7.5 m/s.

Substituting into equation 2

V₂ = [0 (0.02) + 7.5 (0.01) ]/0.03

V₂ = 0.075/0.03

V₂ = 2.5 m/s

But

V = √ (V₁² + V₂²)

Where V = the resultant common velocity, V₁ = Horizontal component of the velocity, V₂ = vertical component of the velocity.

V = √ (5.33²+2.5²)

V = √ (28.41+6.25)

V = √34.66

V = 5.89 m/s

Thus the speed = 5.89 m/s