As part of a carnival game, a 5.00 kg target is freely hanging from a very long and very light wire. Contestants can use one of two 1.5 kg balls to try to hit the target and deflect it high enough to win a prize. Ball A will have an elastic collision and bounce back toward you while ball B will have a nearly perfectly inelastic collision, but rather than sticking to the target, the ball will just drop straight downward to the ground after the collision. You can throw each ball with a velocity of 12 m/s. You are the first to try the game and which ball should you throw? Calculate the expected height the target will reach after each is thrown.

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.

a=d/t

a="triangle"v/t

explanation:

a = acceleration