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2 months ago

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In a carnival game, the player throws a ball at a haystack. For a typical throw, the ball leaves the hay with a speed exactly on

e-half of the entry speed. If the frictional force exerted by the hay is a constant 6.0 N and the haystack is 1.2 m thick, derive an expression for the typical entry speed as a function of the inertia of the ball. Assume horizontal motion only, and ignore any effects due to gravity. What is the typical entry speed if the ball has an inertia of a 0.35 kg?

1 answer:

serg [3.5K]2 months ago

8 0

Réponse:

Ve(m) = √(19.2/m)

Ve(0.35) = 7.407 m/s

Explication:

Données:

- La masse de la balle = m

- La vitesse d'entrée de la balle est Vi = Ve

- La vitesse finale de la balle Vf = 0.5*Ve

- La force de frottement constante sur la balle due à la paille est F = 6 N

- L'épaisseur du tas de paille est s = 1.2 m

- On suppose que le lancer se fait horizontalement et qu'on ignore les effets de la gravité

Objectif:

Déterminer une expression pour la vitesse d'entrée typique en fonction de l'inertie de la balle

Quelle est la vitesse d'entrée typique si la balle a une inertie de 0.35 kg?

Solution:

- Pour connaître la vitesse d'entrée comme fonction de l'inertie, nous allons utiliser la troisième équation du mouvement comme suit:

Vf² = Vi² + 2*a*s

où, a est l'accélération de la balle dans la paille. Nous utiliserons la loi du mouvement de Newton pour cela:

F_net = m*a

La seule force agissant sur la balle en traversant la paille est la force de friction F:

- F = m*a

a = -F/m

- Insérer toutes les quantités dans la troisième équation du mouvement:

(0.5Ve)² = Ve² - 2*F*s / m

0.75Ve² = 2*F*s / m

Ve = √(8*F*s/3*m)

- Remplacer les valeurs:

Ve(m) = √(8*6*1.2/3*m)

Ve(m) = √(19.2/m)

- Pour l'inertie de la balle m = 0.35 kg, la vitesse d'entrée est:

Ve(0.35) = √(19.2/0.35)

Ve(0.35) = 7.407 m/s

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