Problem 1.120 — Perfectly inelastic collision — loss of kinetic energy

Problem Statement

Solve the momentum/collision problem: A bullet of mass $m = 10\,\text{g}$ moving at $v_0 = 800\,\text{m/s}$ embeds in a wooden block of mass $M = 990\,\text{g}$. Find the velocity after collision and the fraction of KE lost. Conservation of momentum: $$mv_0 = (m+M)V$$ $$V = \frac{mv_0}{m+M} = \frac{0.010\times800}{1.000} = \boxed{8.0\,\

Given Information

• $m = 10\,\text{g}$
• $v_0 = 800\,\text{m/s}$
• $M = 990\,\text{g}$

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

$$mv_0 = (m+M)V$$

$$V = \frac{mv_0}{m+M} = \frac{0.010\times800}{1.000} = \boxed{8.0\,\

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

$$\boxed{v_f = \dfrac{m_1 u_1 + m_2 u_2}{m_1+m_2}}$$

Physical Interpretation

In an elastic collision, kinetic energy is perfectly conserved — the objects bounce off like ideal billiard balls. The heavier object always loses a smaller fraction of its momentum than the lighter one.