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

Question

UpStudy AI · Accepted Answer

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.
Option D has the greatest change in momentum of 8 kg·m/s.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Physics Questions