Problem Statement
Solve the momentum/collision problem: Find the dimensions and SI unit of angular momentum. $L = mvr$ (angular momentum = linear momentum × radius) $[L] = ML^2T^{-1}$ Step 1: $L = p \times r = mvr$. Step 2: $[p] = MLT^{-1}$, $[r] = L$. Step 3: $[L] = MLT^{-1} \times L = ML^2T^{-1}$. $$\boxed{[L] = ML^2T^{-1},\quad \text{SI unit} = \text{
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Conservation of linear momentum holds whenever the net external force on a system is zero. In collisions, momentum is always conserved. Additionally, in elastic collisions kinetic energy is also conserved, whereas in perfectly inelastic collisions the objects stick together and kinetic energy is partially converted to heat and deformation.
- $\mathbf{p}_\text{tot} = \sum m_i\mathbf{v}_i = \text{const}$ — conservation of momentum
- Elastic: $\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \text{const}$ — KE conserved
- Inelastic: $m_1v_1 = (m_1+m_2)V$ — perfectly inelastic
- $\eta = \Delta KE/KE_0 = M/(m+M)$ — fractional KE loss (bullet-block)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\boxed{[L] = ML^2T^{-1},\quad \text{SI unit} = \text{
Given Information
- Masses $m_1$, $m_2$ and initial velocities $u_1$, $u_2$ as given
- Type of collision: elastic (KE conserved), perfectly inelastic (objects stick), or partially inelastic
Physical Concepts & Formulas
Linear momentum $\vec{p} = m\vec{v}$ is conserved whenever the net external force on the system is zero. In collisions, the collision forces are internal and huge but brief — the impulse-momentum theorem shows that external forces (gravity, friction) contribute negligible impulse during the short collision time. For elastic collisions, kinetic energy is also conserved, giving two equations for two unknowns. For perfectly inelastic collisions, objects merge and momentum alone governs the outcome.
- $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$ — momentum conservation
- Elastic: $\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$
- Elastic result: $v_1 = \dfrac{(m_1-m_2)u_1+2m_2 u_2}{m_1+m_2}$
- Perfectly inelastic: $(m_1+m_2)v_f = m_1 u_1 + m_2 u_2$
Step-by-Step Solution
Step 1 — Identify type: Elastic, inelastic, or perfectly inelastic.
Step 2 — Write conservation equations:
$$
$$
Step 3 — For elastic collisions, add energy equation or use relative velocity relation: $(u_1 – u_2) = -(v_1-v_2)$.
Step 4 — Solve simultaneously for $v_1$ and $v_2$.
Worked Calculation
Substituting all values with units:
$$
$$
Answer
$$
Answer
$$\boxed{v_f = \dfrac{m_1 u_1 + m_2 u_2}{m_1+m_2}}$$
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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