SMA Physics

Tag: inelastic collisions in one dimensions

  • L11: Inelastic Collisions in One Dimension & Coefficient of Restitution

    Summary

    This is Lecture 11 for the topic “Work, Energy and Power”. You can find all resources related with Work, Energy and Power by clicking on this link. Here we will cover the concept of Inelastic Collisions in one dimension, coefficient of restitution and different aspects of it.

    1. Introduction to Inelastic Collisions

    In the real world, nearly all macroscopic collisions are inelastic. While total linear momentum remains conserved during the impact (due to the absence of external impulsive forces), total mechanical energy (specifically kinetic energy) is not conserved.

    During an inelastic collision:

    • Part of the initial kinetic energy is converted into non-mechanical forms of energy, such as thermal energy, sound energy, and permanent internal deformation.
    • The total momentum before collision equals the total momentum after collision.
    Inelastic Collision in One Dimension

    2. The Coefficient of Restitution ($e$)

    To quantify the “elasticity” or “bounciness” of a collision, physicist Sir Isaac Newton introduced the Coefficient of Restitution ($e$). It is defined as the ratio of the relative velocity of separation to the relative velocity of approach along the line of impact:

    $$e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{v_2 – v_1}{u_1 – u_2}$$

    Where:

    • $u_1, u_2$ are the initial velocities of masses $m_1$ and $m_2$ before collision ($u_1 > u_2$).
    • $v_1, v_2$ are the final velocities after collision ($v_2 > v_1$).

    Ranges and Classifications of $e$:

    1. $e = 1$: Perfectly Elastic Collision (No loss of kinetic energy; relative separation equals relative approach).
    2. $0 < e < 1$: Partially Inelastic Collision (Real-world collisions; kinetic energy is partially lost).
    3. $e = 0$: Perfectly Inelastic Collision (Maximum loss of kinetic energy; the bodies stick together after impact, meaning $v_1 = v_2$).

    3. General Equations for Final Velocities ($v_1$ and $v_2$)

    Consider two bodies of masses $m_1$ and $m_2$ moving with initial velocities $u_1$ and $u_2$.

    By combining the Law of Conservation of Linear Momentum ($m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$) and the Restitution Equation ($v_2 – v_1 = e(u_1 – u_2)$), we can derive the general final velocities for any 1D collision:

    $$v_1 = \frac{(m_1 – em_2)u_1 + (1 + e)m_2 u_2}{m_1 + m_2}$$

    $$v_2 = \frac{(m_2 – em_1)u_2 + (1 + e)m_1 u_1}{m_1 + m_2}$$

    (Note: If you substitute $e = 1$ into these formulas, they seamlessly reduce back to the standard elastic collision equations).

    4. Loss of Kinetic Energy ($\Delta K$) in Inelastic Collisions

    Because kinetic energy is lost during an inelastic collision, $\Delta K = K_{initial} – K_{final}$ is always positive ($>0$).

    Using the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$ and the initial relative velocity $(u_1 – u_2)$, the exact formula for the kinetic energy lost is:

    $$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (1 – e^2)(u_1 – u_2)^2$$

    • If $e = 1$, $\Delta K = 0$ (no energy lost).
    • If $e = 0$, $\Delta K$ is at its maximum value.

    5. Perfectly Inelastic Collisions ($e = 0$)

    When two bodies collide and stick together, $e = 0$, which means $v_1 = v_2 = v_{common}$.

    Applying momentum conservation:

    $$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v_{common}$$

    $$v_{common} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$$

    The loss of kinetic energy in a perfectly inelastic collision simplifies to:

    $$\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (u_1 – u_2)^2$$

    6. Special JEE-Mains Case: Ball Dropping on a Fixed Floor

    A ball is dropped from a height $h_0$ onto a rigid horizontal floor with coefficient of restitution $e$.

    1. Velocity just before first impact: $u = \sqrt{2gh_0}$
    2. Velocity just after first rebound: $v_1 = e u = e\sqrt{2gh_0}$
    3. Height reached after first rebound ($h_1$):
      $$h_1 = \frac{v_1^2}{2g} = e^2 h_0$$
    4. Height after $n$ rebounds ($h_n$):
      $$h_n = e^{2n} h_0$$
    5. Total time elapsed before coming to rest:
      $$T_{total} = \sqrt{\frac{2h_0}{g}} \left( \frac{1 + e}{1 – e} \right)$$
    6. Total distance traveled before coming to rest:
      $$D_{total} = h_0 \left( \frac{1 + e^2}{1 – e^2} \right)$$

    JEE-Mains/NEET Practice Questions

    Question 1: Restitution Calculation

    A ball is moving with a velocity of $10 \text{ m/s}$ towards a stationary wall and rebounds with a velocity of $6 \text{ m/s}$ in the opposite direction. What is the coefficient of restitution ($e$) of the collision?

    (a) $0.4$

    (b) $0.6$

    (c) $0.8$

    (d) $1.0$

    Solution:

    The wall is stationary, so $u_2 = 0$ and $v_2 = 0$.

    Initial velocity of the ball is $u_1 = 10 \text{ m/s}$. Final velocity is $v_1 = -6 \text{ m/s}$ (taking the rebound direction as negative).

    $e = \frac{\text{Relative velocity of separation}}{\text{Relative velocity of approach}} = \frac{0 – (-6)}{10 – 0} = \frac{6}{10} = 0.6$.

    Answer: (b)

    Question 2: Loss of Kinetic Energy

    A block of mass $2 \text{ kg}$ moving at $8 \text{ m/s}$ collides head-on with another stationary block of mass $2 \text{ kg}$. If the coefficient of restitution is $e = 0.5$, how much kinetic energy is lost during the collision?

    (a) $16 \text{ J}$

    (b) $24 \text{ J}$

    (c) $32 \text{ J}$

    (d) $48 \text{ J}$

    Solution:

    Given: $m_1 = 2 \text{ kg}$, $u_1 = 8 \text{ m/s}$, $m_2 = 2 \text{ kg}$, $u_2 = 0$, $e = 0.5$.

    Use the kinetic energy loss formula:

    $\Delta K = \frac{1}{2} \left( \frac{m_1 m_2}{m_1 + m_2} \right) (1 – e^2)(u_1 – u_2)^2$

    $\Delta K = \frac{1}{2} \left( \frac{2 \times 2}{2 + 2} \right) (1 – 0.5^2)(8 – 0)^2$

    $\Delta K = \frac{1}{2} \left( \frac{4}{4} \right) (1 – 0.25)(64)$

    $\Delta K = \frac{1}{2} (1) (0.75)(64) = 0.5 \times 0.75 \times 64 = 24 \text{ J}$.

    Answer: (b)

    Question 3: Rebound Height Progression

    A rubber ball is dropped from a height of $81 \text{ cm}$ onto a floor. If it rebounds to a height of $36 \text{ cm}$, what is the coefficient of restitution? To what height will it rise after the second rebound?

    (a) $e = \frac{2}{3}, h_2 = 16 \text{ cm}$

    (b) $e = \frac{4}{9}, h_2 = 18 \text{ cm}$

    (c) $e = \frac{2}{3}, h_2 = 24 \text{ cm}$

    (d) $e = \frac{1}{2}, h_2 = 9 \text{ cm}$

    Solution:

    1. Find $e$: $h_1 = e^2 h_0 \implies 36 = e^2 (81) \implies e^2 = \frac{36}{81} = \frac{4}{9} \implies e = \frac{2}{3}$.
    2. Find height after second rebound ($h_2$): $h_2 = e^4 h_0$ (or $h_2 = e^2 h_1$).
      $h_2 = \left(\frac{2}{3}\right)^2 \times 36 = \frac{4}{9} \times 36 = 4 \times 4 = 16 \text{ cm}$.
      Answer: (a)

    Question 4: Perfectly Inelastic Impact

    A bullet of mass $m$ is fired horizontally with a velocity $u$ into a stationary wooden block of mass $M$ suspended by a light string. If the bullet embeds itself completely inside the block, what is the fraction of initial kinetic energy converted into heat/internal energy?

    (a) $\frac{M}{m + M}$

    (b) $\frac{m}{m + M}$

    (c) $\frac{m}{M}$

    (d) $\frac{M}{m}$

    Solution:

    1. Initial kinetic energy $K_i = \frac{1}{2}mu^2$.
    2. Final common velocity after embedding ($e = 0$): $v = \frac{mu}{m + M}$.
    3. Final kinetic energy $K_f = \frac{1}{2}(m + M)v^2 = \frac{1}{2}(m + M)\left(\frac{mu}{m + M}\right)^2 = \frac{1}{2}mu^2 \left(\frac{m}{m + M}\right)$.
    4. Energy lost (converted to heat): $\Delta K = K_i – K_f = \frac{1}{2}mu^2 \left(1 – \frac{m}{m + M}\right) = \frac{1}{2}mu^2 \left(\frac{M}{m + M}\right)$.
    5. Fraction lost = $\frac{\Delta K}{K_i} = \frac{M}{m + M}$.
      Answer: (a)