Question

A steel ball with mass 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 g = 9.8 m/s² .

Answer 1

We can calculate the impulse J:

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

Step-by-step explanation:

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

• 
  `\[ J = \Delta p \]`

where
`\( \Delta p \)`is the change in momentum.

The change in momentum
`(\( \Delta p \))` can be calculated using the following formula:

• 
  `\[ \Delta p = m \cdot \Delta v \]`

where:

• m is the mass of the object,
• 
  `\( \Delta v \)` is the change in velocity.

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

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

where:

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

Substituting in the values:

• 
  `\[ v_f^2 = 0 + 2 \cdot 9.8 \cdot 1.69 \]`
• 
  `\[ v_f = √(2 \cdot 9.8 \cdot 1.69) \]`

Now that we have the final velocity, we can calculate the change in velocity
`(\( \Delta v \))`:

• 
  `\[ \Delta v = v_f - v_i \]`
• 
  `\[ \Delta v = √(2 \cdot 9.8 \cdot 1.69) - 0 \]`

Now, we can calculate the impulse J:

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

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