A rubber ball of mass 250 g hits a wall normally with a velocity of 10 m s-1 and bounces back with a velocity of 8 m s-1. The impulse is _____ N s.

1. Break Down the Problem

To find the impulse, we need to analyze the change in momentum of the rubber ball as it hits the wall and bounces back.

2. Relevant Concepts

• The impulse experienced by an object is equal to the change in momentum: $\text{Impulse} = \Delta p = p_{\text{final}} - p_{\text{initial}}$
• Momentum ($p$) is calculated using the formula: $p = m \cdot v$ where $m$ is the mass and $v$ is the velocity.

3. Analysis and Detail

1. Convert mass from grams to kilograms: $m = 250 \, \text{g} = 0.250 \, \text{kg}$

2. Calculate initial momentum ($p_{\text{initial}}$):

   • The initial velocity is $v_{\text{initial}} = 10 \, \text{m/s}$ towards the wall (we can consider this as negative since it's hitting the wall). $p_{\text{initial}} = m \cdot v_{\text{initial}} = 0.250 \, \text{kg} \cdot (-10 \, \text{m/s}) = -2.5 \, \text{kg m/s}$

3. Calculate final momentum ($p_{\text{final}}$):

   • The final velocity is $v_{\text{final}} = 8 \, \text{m/s}$ away from the wall (positive). $p_{\text{final}} = m \cdot v_{\text{final}} = 0.250 \, \text{kg} \cdot 8 \, \text{m/s} = 2.0 \, \text{kg m/s}$

4. Calculate the change in momentum ($\Delta p$): $\Delta p = p_{\text{final}} - p_{\text{initial}} = 2.0 \, \text{kg m/s} - (-2.5 \, \text{kg m/s}) = 2.0 + 2.5 = 4.5 \, \text{kg m/s}$

4. Verify and Summarize

• Initial momentum: $-2.5 \, \text{kg m/s}$
• Final momentum: $2.0 \, \text{kg m/s}$
• Change in momentum (impulse): $4.5 \, \text{kg m/s} = 4.5 \, \text{Ns}$

Final Answer

The impulse is $4.5 \, \text{N s}$.