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natali 33 [55]
3 years ago
14

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.
Physics
1 answer:
Nina [5.8K]3 years ago
6 0

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

<u>Throwing ball A</u>; 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

<u>Throwing ball B;</u> 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.

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Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

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20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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