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

Response:

When ball A is thrown, the target's height will be 2.65 m

The height achieved by the target when ball A is used is 2.2 m

Using ball A results in the target reaching a higher altitude, so I will opt to throw ball A

Clarification:

Provided;

mass of the target, m₁ = 5.00 kg

mass of the ball, m₂ = 1.5 kg

initial velocity of the throws, u = 12 m/s

Throwing ball A; we use the conservation of linear momentum principle for elastic collisions;

m₁u₁ + m₂u₂ = v₁m₁ + v₂m₂

The target's initial velocity, u₁ = 0

The ball rebounded at the same speed, v₂ = -12 m/s

The post-collision speed of the target is 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 height the target reaches can be calculated with;

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; utilize the conservation of energy for the inelastic collision;

the ball's kinetic energy transforms into 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 cause the target to achieve a higher elevation, therefore, I will choose to throw ball A.