Answer:
Height ball A will deflect the target is 2.65 m
Height ball A will deflectthe target is 2.2 m
Ball A, will deflect the target to greater height, thus i will throw ball A
Explanation:
Given;
mass of the target, m₁ = 5.00 kg
mass of the ball, m₂ = 1.5 kg
initial velocity of each throw, u = 12 m/s
Throwing ball A; apply the principle of conservation linear momentum for elastic collision;
m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂
The initial velocity of the target, u₁ = 0
The ball bounced back at the same speed, v₂ = -12 m/s
the velocity of the target after collision, = v₁
0 + 1.5 x 12 = v₁(5) + (-12 x 1.5)
18 = 5v₁ - 18
18 + 18 = 5v₁
36 = 5v₁
v₁ = 36/5
v₁ = 7.2 m/s
The vertical height reached by the target is given by;
v₁² = u₁² + 2gh
v₁² = 0 + 2gh
v₁² = 2gh
h = v₁² / 2g
h = (7.2)² / (2 x 9.8)
h = 2.65 m
Throwing ball B; apply the principle of conservation energy for the inelastic collision;
the kinetic energy of the ball will be converted to the potential energy of the target.
¹/₂m₁u₂² = mgh
¹/₂(1.5)(12)² = (5 x 9.8)h
108 = 49h
h = 108 / 49
h = 2.2 m
Ball A will deflect the target to greater height, thus i will throw ball A.
