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A 0,34-kg volleyball is flying west at 1,98 m/s when it strikes a stationary 0,60-kg basketball dead centre. The volleyball rebounds east at 0,79 m/s. What will be the velocity of the basketball immediately after impact? Choose the east direction as positive.

Respuesta :

maforchrist

Explanation:

use the law of conservation of momentum

M1U1+M2U2=M1V1+M2V2

m-mass

u-initial velocity

v-final velocity

1-volleyball

2-basket ball dead center

so using the above formula

0.34×1.98+0.64×0=0.34×0.79+0.64×V2

to get

0.6732+0 = 0.2686+0.64V2

0.4046=0.64V2

divide both sides by 0.64 to get V2 which is the final velocity of basketball

0.6322m/s=V2

onyebuchinnaji

The final velocity of the basketball after the impact is 1.643 m/s west.

The given parameters;

  • mass of the volleyball, m = 0.34 kg
  • initial velocity of the volleyball, u₁ = 1.98 m/s
  • mass of the stationary basketball, m₂ = 0.6 kg
  • final velocity of the volleyball, v₁ = 0.79 m/s

The final velocity of the basketball after the impact is calculated by applying the principle of conservation of linear momentum;

m₁u₁  +  m₂u₂  =  m₁v₁   +  m₂v₂

0.34(-1.98) + 0.6(0) = 0.79(0.34) + 0.6(v₂)

-0.673 = 0.316 + 0.6v₂

-0.989 = 0.6v₂

[tex]v_2 = \frac{- 0.989}{0.6} \\\\v_2 = -1.643 \ m/s\\\\v_2 = 1.643 \ m/s \ west[/tex]

Thus, the final velocity of the basketball after the impact is 1.643 m/s west.

Learn more here:https://brainly.com/question/24424291

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