In which one of the following is the change in momentum of the objcct the greatest Let the mass of the object be 2 kg A Initial momentum \( =3 \mathrm{~kg} \mathrm{~m} \mathrm{~s}{ }^{-1} \) dowwards final momentum \( =21 \mathrm{~g} \mathrm{~m} \mathrm{~s}^{-1} \) upwands 3 Initial velocity \( =2 \mathrm{~ms}^{-1} \) southwards, final velociry \( =1 \mathrm{~ms}^{-1} \) northwands E Initial velocity \( =2 \mathrm{~m}^{-1} \) towands the righr, final velocify \( =1 \mathrm{~ms}^{-1} \) rowands the nght D. Initial momentum \( =4 \mathrm{kgms}^{-1} \) dowwwants, final momenturn \( =4 \mathrm{~kg} \mathrm{~ms}^{-1} \) upwands

To determine which scenario results in the greatest change in momentum for a 2 kg object, let's analyze each option step by step. ### Change in Momentum (\(\Delta p\)) Calculation The change in momentum is calculated using the formula: \[ \Delta p = |p_{\text{final}} - p_{\text{initial}}| \] where \( p = mv \) (momentum). ### Option A - **Initial Momentum (\(p_{\text{initial}}\))**: \( 3 \, \text{kg} \cdot \text{m/s} \) downwards - **Final Momentum (\(p_{\text{final}}\))**: \( 21 \, \text{g} \cdot \text{m/s} = 0.021 \, \text{kg} \cdot \text{m/s} \) upwards \[ \Delta p = |0.021 \, \text{kg} \cdot \text{m/s} - 3 \, \text{kg} \cdot \text{m/s}| = | -2.979 \, \text{kg} \cdot \text{m/s} | = 2.979 \, \text{kg} \cdot \text{m/s} \] ### Option B - **Initial Velocity**: \( 2 \, \text{m/s} \) southwards - **Final Velocity**: \( 1 \, \text{m/s} \) northwards - **Mass**: \( 2 \, \text{kg} \) \[ p_{\text{initial}} = 2 \, \text{kg} \times 2 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s} \, \text{south} \] \[ p_{\text{final}} = 2 \, \text{kg} \times 1 \, \text{m/s} = 2 \, \text{kg} \cdot \text{m/s} \, \text{north} \] \[ \Delta p = |2 - (-4)| = 6 \, \text{kg} \cdot \text{m/s} \] ### Option E - **Initial Velocity**: \( 2 \, \text{m/s} \) to the right - **Final Velocity**: \( 1 \, \text{m/s} \) to the left - **Mass**: \( 2 \, \text{kg} \) \[ p_{\text{initial}} = 2 \, \text{kg} \times 2 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s} \, \text{right} \] \[ p_{\text{final}} = 2 \, \text{kg} \times 1 \, \text{m/s} = 2 \, \text{kg} \cdot \text{m/s} \, \text{left} = -2 \, \text{kg} \cdot \text{m/s} \] \[ \Delta p = |-2 - 4| = 6 \, \text{kg} \cdot \text{m/s} \] ### Option D - **Initial Momentum (\(p_{\text{initial}}\))**: \( 4 \, \text{kg} \cdot \text{m/s} \) downwards - **Final Momentum (\(p_{\text{final}}\))**: \( 4 \, \text{kg} \cdot \text{m/s} \) upwards \[ \Delta p = |4 \, \text{kg} \cdot \text{m/s} - (-4 \, \text{kg} \cdot \text{m/s})| = 8 \, \text{kg} \cdot \text{m/s} \] ### Conclusion - **Option D** results in the **greatest change in momentum** of **8 kg·m/s**. **Answer:** Option D. The change in momentum is greatest when momentum changes from 4 kg·m/s downward to 4 kg·m/s upward.