Question

When a ball of mass 2.2 kg collides with a wall with a speed of 10 m/s and rebounds with a speed of 8 m/s. Find impulse.

Hint:

A common trap is subtracting the speeds directly: $\Delta p \neq m(|v_i| - |v_f|)$. Because it rebounds, the velocities are in opposite directions, so you must effectively add their magnitudes: $J = m(v_{initial} - (-v_{final})) = m(v_i + v_f)$.

Answer 1

Step 1: Understanding the Concept:
Impulse (\(J\)) is defined as the force integrated over time, which according to the impulse-momentum theorem, is exactly equal to the change in momentum (\(\Delta p\)) of the object. Momentum is a vector quantity, so direction matters critically.
Step 2: Key Formula or Approach:
\[ \text{Impulse } \vec{J} = \Delta\vec{p} = \vec{p}_f - \vec{p}_i = m\vec{v}_f - m\vec{v}_i \]
Define a coordinate system. Let the direction towards the wall be positive.
Step 3: Detailed Explanation:
Given values:
Mass of ball, \(m = 2.2\) kg
Initial speed towards wall, \(|v_i| = 10\) m/s
Final speed rebounding from wall, \(|v_f| = 8\) m/s
Let's assign vector signs based on our coordinate system (towards wall = +):
Initial velocity vector, \(\vec{v}_i = +10\) m/s
Final velocity vector, \(\vec{v}_f = -8\) m/s (it reversed direction)
Now, calculate the initial and final momentum:
Initial momentum, \(\vec{p}_i = m \cdot \vec{v}_i = 2.2 \times (+10) = +22\) kg m/s
Final momentum, \(\vec{p}_f = m \cdot \vec{v}_f = 2.2 \times (-8) = -17.6\) kg m/s
Calculate Impulse:
\[ \vec{J} = \vec{p}_f - \vec{p}_i \]
\[ \vec{J} = -17.6 - 22 \]
\[ \vec{J} = -39.6 \text{ kg m/s} \text{ (or N s)} \]
The negative sign indicates the impulse is directed away from the wall, which makes sense as the wall exerts force outward to stop and reverse the ball.
Usually, questions ask for the magnitude of the impulse.
Magnitude of Impulse = \(|-39.6| = 39.6\) N s.
Step 4: Final Answer:
The magnitude of the impulse is 39.6 N s.