Answer:
5.89 m/s
Step-by-step 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