2. In which one of the following is the change in momenrum of the object the grearest? Ler the mass of the object be 2 kg A Initial momentum $$ =3 \mathrm{~kg} \mathrm{~ms}{ }^{-1} $$ dowwards, final momentum $$ =2 \mathrm{~kg} \mathrm{~ms}^{-1} $$ upwards 3 Initial velocity $$ =2 \mathrm{~ms}^{-1} $$ southwards, final velociry $$ =1 \mathrm{~ms}^{-1} $$ northwands C Initial velocity $$ =2 \mathrm{~m}-s^{-1} $$ towards the righe, final veloary $$ =1 \mathrm{~ms}^{-1} $$ towards the night D Initial momenturn $$ =4 \mathrm{kgms} s^{-1} $$ downwands, final momentum $$ =4 \mathrm{~kg} \mathrm{~m}^{-1} $$ upwards

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To determine which scenario has the greatest change in momentum, we need to calculate the change in momentum for each option. The change in momentum ($$\Delta p$$) can be calculated using the formula:

$$
\Delta p = p_{\text{final}} - p_{\text{initial}}
$$

where $$p$$ is the momentum.

Let's analyze each option step by step:

### Option A
- Initial momentum: $$p_{\text{initial}} = 3 \, \text{kg m/s}$$ (downwards)
- Final momentum: $$p_{\text{final}} = 2 \, \text{kg m/s}$$ (upwards)

Calculating the change in momentum:
$$
\Delta p_A = p_{\text{final}} - p_{\text{initial}} = 2 - 3 = -1 \, \text{kg m/s}
$$

### Option B
- Initial velocity: $$v_{\text{initial}} = 2 \, \text{m/s}$$ (southwards)
- Final velocity: $$v_{\text{final}} = 1 \, \text{m/s}$$ (northwards)

Calculating the initial and final momentum:
$$
p_{\text{initial}} = m \cdot v_{\text{initial}} = 2 \cdot 2 = 4 \, \text{kg m/s} \, \text{(southwards)}
$$
$$
p_{\text{final}} = m \cdot v_{\text{final}} = 2 \cdot 1 = 2 \, \text{kg m/s} \, \text{(northwards)}
$$

Calculating the change in momentum:
$$
\Delta p_B = p_{\text{final}} - p_{\text{initial}} = 2 - 4 = -2 \, \text{kg m/s}
$$

### Option C
- Initial velocity: $$v_{\text{initial}} = 2 \, \text{m/s}$$ (towards the right)
- Final velocity: $$v_{\text{final}} = 1 \, \text{m/s}$$ (towards the left)

Calculating the initial and final momentum:
$$
p_{\text{initial}} = m \cdot v_{\text{initial}} = 2 \cdot 2 = 4 \, \text{kg m/s} \, \text{(right)}
$$
$$
p_{\text{final}} = m \cdot v_{\text{final}} = 2 \cdot (-1) = -2 \, \text{kg m/s} \, \text{(left)}
$$

Calculating the change in momentum:
$$
\Delta p_C = p_{\text{final}} - p_{\text{initial}} = -2 - 4 = -6 \, \text{kg m/s}
$$

### Option D
- Initial momentum: $$p_{\text{initial}} = 4 \, \text{kg m/s}$$ (downwards)
- Final momentum: $$p_{\text{final}} = 4 \, \text{kg m/s}$$ (upwards)

Calculating the change in momentum:
$$
\Delta p_D = p_{\text{final}} - p_{\text{initial}} = 4 - 4 = 0 \, \text{kg m/s}
$$

### Summary of Changes in Momentum
- $$\Delta p_A = -1 \, \text{kg m/s}$$
- $$\Delta p_B = -2 \, \text{kg m/s}$$
- $$\Delta p_C = -6 \, \text{kg m/s}$$
- $$\Delta p_D = 0 \, \text{kg m/s}$$

### Conclusion
The greatest change in momentum occurs in **Option A** with a change of $$-1 \, \text{kg m/s}$$.
The greatest change in momentum is in Option A, with a change of $$-1 \, \text{kg m/s}$$.

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