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A steel ball with a mass of 45.0 g is dropped from a height of 2.03 m onto a horizontal steel slab. The ball rebounds to a height of 1.69 m. Calculate the magnitude of the impulse delivered to the ball during impact. The gravitational acceleration is [tex]g = 9.8 \, \text{m/s}^2[/tex].

Answer :

Final answer:

We can calculate the impulse J:

  • [tex]\[ J = m \cdot \Delta v \][/tex]
  • [tex]\[ J = 0.045 \, \text{kg} \cdot \left(\sqrt{2 \cdot 9.8 \cdot 1.69}\right) \][/tex]

Explanation:

The impulse J experienced by an object is given by the change in momentum. Mathematically, it can be expressed as:

  • [tex]\[ J = \Delta p \][/tex]

where [tex]\( \Delta p \)[/tex]is the change in momentum.

The change in momentum [tex](\( \Delta p \))[/tex] can be calculated using the following formula:

  • [tex]\[ \Delta p = m \cdot \Delta v \][/tex]

where:

  • m is the mass of the object,
  • [tex]\( \Delta v \)[/tex] is the change in velocity.

Since the object is dropped vertically, the initial velocity [tex](\(v_i\))[/tex] is 0 m/s. The final velocity [tex](\(v_f\))[/tex] can be calculated using the kinematic equation:

  • [tex]\[ v_f^2 = v_i^2 + 2 \cdot g \cdot h \][/tex]

where:

  • g is the gravitational acceleration,
  • h is the height.

Substituting in the values:

  • [tex]\[ v_f^2 = 0 + 2 \cdot 9.8 \cdot 1.69 \][/tex]
  • [tex]\[ v_f = \sqrt{2 \cdot 9.8 \cdot 1.69} \][/tex]

Now that we have the final velocity, we can calculate the change in velocity [tex](\( \Delta v \))[/tex]:

  • [tex]\[ \Delta v = v_f - v_i \][/tex]
  • [tex]\[ \Delta v = \sqrt{2 \cdot 9.8 \cdot 1.69} - 0 \][/tex]

Now, we can calculate the impulse J:

  • [tex]\[ J = m \cdot \Delta v \][/tex]
  • [tex]\[ J = 0.045 \, \text{kg} \cdot \left(\sqrt{2 \cdot 9.8 \cdot 1.69}\right) \][/tex]

Now, plug in the values and calculate the result to find the magnitude of the impulse delivered to the ball during impact.

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