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A rubber ball of mass 250 g hits a wall normally with a velocity of 10 m s-1 and bounces back with a velocity of 8 m s-1. The impulse is _____ N s.Select an answerA– 0.5B+ 0.5C– 4.5D+ 4.5

Question

A rubber ball of mass 250 g hits a wall normally with a velocity of 10 m s-1 and bounces back with a velocity of 8 m s-1. The impulse is _____ N s.Select an answerA– 0.5B+ 0.5C– 4.5D+ 4.5

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Solution

Primero, recordemos que el impulso se define como el cambio en la cantidad de movimiento (o momento lineal) de un objeto. La fórmula para el impulso es:

Impulso=Δp=mΔv \text{Impulso} = \Delta p = m \Delta v

donde m m es la masa del objeto y Δv \Delta v es el cambio en la velocidad.

Dado:

  • Masa del balón, m=250 g=0.25 kg m = 250 \, \text{g} = 0.25 \, \text{kg} (convertimos gramos a kilogramos)
  • Velocidad inicial, vi=10 m/s v_i = 10 \, \text{m/s}
  • Velocidad final, vf=−8 m/s v_f = -8 \, \text{m/s} (es negativa porque rebota en la dirección opuesta)

El cambio en la velocidad, Δv \Delta v , es:

Δv=vf−vi=−8 m/s−10 m/s=−18 m/s \Delta v = v_f - v_i = -8 \, \text{m/s} - 10 \, \text{m/s} = -18 \, \text{m/s}

Ahora, calculamos el impulso:

Impulso=mΔv=0.25 kg×(−18 m/s)=−4.5 N\cdotps \text{Impulso} = m \Delta v = 0.25 \, \text{kg} \times (-18 \, \text{m/s}) = -4.5 \, \text{N·s}

Por lo tanto, la respuesta correcta es:

C – 4.5

This problem has been solved

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