Problem Statement
Solve the momentum/collision problem: Find the dimensions and SI unit of linear momentum. $p = mv$ $[p] = MLT^{-1}$ Step 1: $p = mv$, $[m] = M$, $[v] = LT^{-1}$. Step 2: $[p] = M \cdot LT^{-1} = MLT^{-1}$. $$\boxed{[p] = MLT^{-1},\quad \text{SI unit} = \text{kg m s}^{-1}}$$
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Conservation of linear momentum holds whenever the net external force on a system is zero. In collisions, momentum is always conserved. Additionally, in elastic collisions kinetic energy is also conserved, whereas in perfectly inelastic collisions the objects stick together and kinetic energy is partially converted to heat and deformation.
- $\mathbf{p}_\text{tot} = \sum m_i\mathbf{v}_i = \text{const}$ — conservation of momentum
- Elastic: $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \text{const}$ — KE conserved
- Inelastic: $m_1v_1 = (m_1+m_2)V$ — perfectly inelastic
- $\eta = \Delta KE/KE_0 = M/(m+M)$ — fractional KE loss (bullet-block)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{[p] = MLT^{-1},\quad \text{SI unit} = \text{kg m s}^{-1}}$$
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$$
$$v_f = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}\quad\text{(perfectly inelastic)}$$
Answer
$$\boxed{[p] = MLT^{-1},\quad \text{SI unit} = \text{kg m s}^{-1}}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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