Inelastic Collisions — Revision Notes

⚡ 30-Second Revision

Linear Momentum (p): — $\vec{p} = m\vec{v}$. Conserved in all collisions (isolated system). \n- Kinetic Energy (KE): $KE = \frac{1}{2}mv^2$. NOT conserved in inelastic collisions. \n- Inelastic Collision: Momentum conserved, KE NOT conserved. $0 \le e < 1$. \n- Perfectly Inelastic Collision: Objects stick together. $e=0$. Maximum KE loss. \n- Conservation of Momentum (1D): $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. \n- Coefficient of Restitution (e): $e = \frac{|v_2 - v_1|}{|u_1 - u_2|}$. \n- Common Final Velocity (Perfectly Inelastic): $V = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$. \n- Kinetic Energy Loss (General): $\Delta KE = KE_{initial} - KE_{final}$. \n- Max KE Loss (Perfectly Inelastic): $\Delta KE_{max} = \frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$.

2-Minute Revision

Inelastic collisions are interactions where the total linear momentum of the system is always conserved, but the total kinetic energy is not. This means some of the initial kinetic energy is converted into other forms like heat, sound, or deformation.

The degree of 'bounciness' is quantified by the coefficient of restitution, 'e', which for inelastic collisions ranges from $0 \le e < 1$. \n\nA crucial type is the perfectly inelastic collision, where objects stick together after impact and move with a common final velocity.

In this case, $e=0$, and the kinetic energy loss is maximum. To solve problems, always start with the conservation of linear momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. If 'e' is given, use $v_2 - v_1 = e(u_1 - u_2)$ as a second equation.

For energy loss, calculate initial and final kinetic energies and find their difference. Remember to use consistent sign conventions for velocities and convert units to SI.

5-Minute Revision

Inelastic collisions are a core concept in mechanics, characterized by the conservation of linear momentum but the non-conservation of kinetic energy. When objects collide inelastically, a portion of their initial kinetic energy is transformed into other forms, such as heat, sound, or internal energy causing permanent deformation.

This energy dissipation is what defines an inelastic collision. \n\nKey Equations and Concepts: \n1. Conservation of Linear Momentum: This is the bedrock principle. For any collision in an isolated system, the total momentum before equals the total momentum after.

For a 1D collision: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. Remember to assign positive and negative signs for velocities based on direction. \n2. Coefficient of Restitution (e): This value indicates the elasticity of the collision.

For inelastic collisions, $0 \le e < 1$. It's defined as $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{|v_2 - v_1|}{|u_1 - u_2|}$. This equation provides a second relation between velocities.

\n3. Perfectly Inelastic Collisions: This is a special case where $e=0$. The objects stick together after impact and move with a common final velocity, $V$. The formula for $V$ is derived directly from momentum conservation: $V = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$.

This type of collision results in the maximum possible kinetic energy loss consistent with momentum conservation. \n4. Kinetic Energy Loss: The energy lost is $\Delta KE = KE_{initial} - KE_{final}$.

For a perfectly inelastic collision, the maximum energy loss is given by $\Delta KE_{max} = \frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$. This formula is very efficient for calculating energy loss in such scenarios.

\n\nWorked Example: A $0.02,\text{kg}$ bullet moving at $500,\text{m/s}$ strikes a $1.98,\text{kg}$ block at rest and gets embedded. What is the common velocity and the kinetic energy lost? \nSolution: \n* Common Velocity (V): $m_b u_b + m_w u_w = (m_b + m_w)V$.

\n $(0.02)(500) + (1.98)(0) = (0.02 + 1.98)V$. \n $10 = (2.00)V \implies V = 5,\text{m/s}$. \n* Kinetic Energy Lost: \n $KE_{initial} = \frac{1}{2}m_b u_b^2 = \frac{1}{2}(0.02)(500)^2 = 0.01 \times 250000 = 2500,\text{J}$.

\n $KE_{final} = \frac{1}{2}(m_b + m_w)V^2 = \frac{1}{2}(2.00)(5)^2 = 1 \times 25 = 25,\text{J}$. \n $\Delta KE = KE_{initial} - KE_{final} = 2500 - 25 = 2475,\text{J}$. \n (Alternatively, using the formula: $\Delta KE_{max} = \frac{1}{2}\frac{(0.

02)(1.98)}{0.02+1.98}(500-0)^2 = \frac{1}{2}\frac{0.0396}{2}(250000) = 0.0099 \times 250000 = 2475, ext{J}$). \n\nMastering these concepts and formulas, along with careful attention to signs and units, will ensure success in NEET problems on inelastic collisions.

Prelims Revision Notes

For NEET, inelastic collisions are critical. Remember these key points: \n\n1. Momentum Conservation is Universal: In any collision (elastic or inelastic), for an isolated system, total linear momentum is conserved.

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$. Always use vector notation or consistent sign conventions for 1D problems. \n2. Kinetic Energy is NOT Conserved: This is the defining characteristic of an inelastic collision.

Some initial kinetic energy is converted into heat, sound, or deformation. Therefore, $KE_{initial} > KE_{final}$. \n3. Coefficient of Restitution (e): This parameter quantifies the degree of inelasticity.

For inelastic collisions, $0 \le e < 1$. It's defined as $e = \frac{\text{relative speed of separation}}{\text{relative speed of approach}} = \frac{|v_2 - v_1|}{|u_1 - u_2|}$. \n4. **Perfectly Inelastic Collision ($e=0$):** This is a very common NEET scenario.

Objects stick together after impact, moving with a common final velocity $V = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$. The kinetic energy loss is maximum in this case. \n5. Calculating Energy Loss: \n * General method: Calculate $KE_{initial}$ and $KE_{final}$, then $\Delta KE = KE_{initial} - KE_{final}$.

\n * Shortcut for perfectly inelastic: $\Delta KE_{max} = \frac{1}{2}\frac{m_1m_2}{m_1+m_2}(u_1-u_2)^2$. This formula is highly efficient. \n6. Bullet-Block Problems: These are two-step problems. First, apply momentum conservation during the perfectly inelastic collision.

Second, apply conservation of mechanical energy (or kinematics) to the subsequent motion of the combined mass (e.g., swinging upwards against gravity). \n7. Impulse: Recall that impulse is the change in momentum ($J = \Delta p = m\Delta v$).

It's a vector quantity. \n8. Units: Always work in SI units (kg, m/s, J, N\cdot s). Be careful with gram to kilogram conversion. \n\nPractice applying these principles to various scenarios, especially those involving objects moving in opposite directions, and problems where one object is initially at rest.

Vyyuha Quick Recall

In Momentum Conserved, Kinetic Energy Lost (IMCKEL) for Inelastic Collisions. \nEquals Zero Sticks Together (EZST) for Perfectly Inelastic.