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You drop a 14-g ball from a height of 1.5 m and it only bounces back to a height of 0.85 m. what was the total impulse on the ball when it hit the floor? (ignore air resistance.)

Respuesta :

shinmin
Remember that  impulse = change in momentum 

this means we compute the momentum of the ball just before impression and just after; we know the mass, so we find the speeds 

the ball falls for 1.5m and will achieve a speed given by energy conservation: 

1/2 mv^2 = mgh => v=sqrt[2gh]=5.42m/s 

since it rises only to 0.85 m, we compute the initial speed after power from the same equation and get 
v(after)=sqrt[2*9.81m/s/s*0.85m] = 4.0837 m... 

now, recall that momentum is a vector, so that the momentum down has one sign and the momentum up has a positive sign, so we have 

impulse = delta (mv) = m delta v = 0.014 kx (4.08m/s - (-5.42m/s) = 0.133 kgm/s 
onyebuchinnaji

The total impulse on the ball when it hit the floor is 0.05 Ns.

The given parameters;

  • mass of the ball, m = 14 g = 0.014 kg
  • initial height of the ball, h = 1.5 m
  • final height of the ball, h₂ = 0.85 m

Apply the principle of conservation of energy as shown below;

ΔK.E = ΔP.E

[tex]\frac{1}{2} m\Delta v^2 = mg\Delta h\\\\\Delta v^2 = 2g\Delta h\\\\\Delta v = \sqrt{2g\Delta h} \\\\\Delta v = \sqrt{2\times 9.8 \times (1.5-0.85)} \\\\\Delta v = 3.57 \ m/s[/tex]

The impulse experienced by the ball is the change in the momentum of the ball;

J = ΔP = mΔv

J = 0.014 x 3.57

J = 0.05 N.s

Thus, the total impulse on the ball when it hit the floor is 0.05 Ns.

Learn more here:https://brainly.com/question/23927723

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