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A player caught a cricket ball of mass 150 g150 g moving at a speed of 20 m/s20 m/s. If the catching process is completed in 0.1 s0.1 s, the magnitude of force exerted by the ball on the hand of the player is:

Question

A player caught a cricket ball of mass 150 g150 g moving at a speed of 20 m/s20 m/s. If the catching process is completed in 0.1 s0.1 s, the magnitude of force exerted by the ball on the hand of the player is:

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Solution

Primero, convertimos la masa de la bola de cricket de gramos a kilogramos: 150g=0.150kg 150 \, \text{g} = 0.150 \, \text{kg}

Luego, utilizamos la segunda ley de Newton, que establece que la fuerza es igual al cambio en el momento (impulso) dividido por el tiempo durante el cual ocurre el cambio: F=ΔpΔt F = \frac{\Delta p}{\Delta t}

El momento inicial (pi p_i ) de la bola es: pi=mv=0.150kg20m/s=3kgm/s p_i = m \cdot v = 0.150 \, \text{kg} \cdot 20 \, \text{m/s} = 3 \, \text{kg} \cdot \text{m/s}

El momento final (pf p_f ) de la bola es cero, ya que la bola se detiene: pf=0 p_f = 0

El cambio en el momento (Δp \Delta p ) es: Δp=pfpi=03kgm/s=3kgm/s \Delta p = p_f - p_i = 0 - 3 \, \text{kg} \cdot \text{m/s} = -3 \, \text{kg} \cdot \text{m/s}

El tiempo durante el cual ocurre el cambio (Δt \Delta t ) es: Δt=0.1s \Delta t = 0.1 \, \text{s}

Ahora, calculamos la fuerza: F=ΔpΔt=3kgm/s0.1s=30N F = \frac{\Delta p}{\Delta t} = \frac{-3 \, \text{kg} \cdot \text{m/s}}{0.1 \, \text{s}} = -30 \, \text{N}

La magnitud de la fuerza es: F=30N |F| = 30 \, \text{N}

Por lo tanto, la magnitud de la fuerza ejercida por la bola en la mano del jugador es: 30N 30 \, \text{N}

This problem has been solved

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