Elastic Collisions — Revision Notes

⚡ 30-Second Revision

Definition: — Total linear momentum and total kinetic energy are conserved.
Coefficient of Restitution: — $e=1$.
1D Momentum Conservation: — $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
1D Kinetic Energy Conservation: — $rac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$
Relative Velocity Relation: — $u_1 - u_2 = v_2 - v_1$ (relative speed of approach = relative speed of separation)
Final Velocities (General 1D):

- v_1 = left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + left(\frac{2m_2}{m_1 + m_2}\right)u_2 - v_2 = left(\frac{2m_1}{m_1 + m_2}\right)u_1 + left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2

Special Case ($m_1=m_2, u_2=0$): — $v_1=0, v_2=u_1$ (velocities exchange)
Special Case ($m_1 ll m_2, u_2=0$): — $v_1 approx -u_1, v_2 approx 0$ (light body bounces back, heavy body at rest)
Special Case ($m_1 gg m_2, u_2=0$): — $v_1 approx u_1, v_2 approx 2u_1$ (heavy body continues, light body moves with double speed)

2-Minute Revision

Elastic collisions are ideal interactions where both linear momentum and kinetic energy are conserved. This means the total 'oomph' and total 'energy of motion' of the system remain unchanged before and after the collision.

The coefficient of restitution, 'e', for an elastic collision is always 1, signifying perfect rebound without energy loss. A key property for one-dimensional elastic collisions is that the relative speed at which objects approach each other before impact is equal to the relative speed at which they separate after impact ($u_1 - u_2 = v_2 - v_1$).

This relation, combined with momentum conservation, allows us to derive formulas for final velocities. Remember important special cases: if two equal masses collide elastically and one is at rest, they exchange velocities.

If a light object hits a massive stationary object, the light object bounces back with nearly the same speed, and the massive object remains almost at rest. Conversely, if a massive object hits a light stationary object, the massive object continues almost unaffected, and the light object shoots forward with nearly double the massive object's initial speed.

Always pay attention to directions using proper sign conventions for velocities.

5-Minute Revision

Elastic collisions are fundamental in physics, characterized by the simultaneous conservation of two critical quantities: total linear momentum and total kinetic energy. For a system of two colliding bodies, $m_1$ and $m_2$, with initial velocities $u_1$ and $u_2$, and final velocities $v_1$ and $v_2$:

1
Conservation of Linear Momentum: — $m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
2
Conservation of Kinetic Energy: — $rac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2$

These two equations are sufficient to solve for the two unknown final velocities in a one-dimensional collision. A powerful shortcut derived from these is the relative velocity relation: $u_1 - u_2 = v_2 - v_1$. This means the relative speed of approach equals the relative speed of separation. This also implies the coefficient of restitution ($e$) is 1 for elastic collisions.

General Formulas for Final Velocities (1D):

v_1 = left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + left(\frac{2m_2}{m_1 + m_2}\right)u_2 v_2 = left(\frac{2m_1}{m_1 + m_2}\right)u_1 + left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2

Crucial Special Cases (for NEET):

Equal Masses ($m_1 = m_2 = m$) and $u_2 = 0$: — $v_1 = 0$, $v_2 = u_1$. (Velocities exchange)

* *Example:* A $5,\text{kg}$ ball at $10,\text{m/s}$ hits a stationary $5,\text{kg}$ ball. The first ball stops, the second moves at $10,\text{m/s}$.

Light Body with Massive Body at Rest ($m_1 ll m_2$, $u_2 = 0$): — $v_1 approx -u_1$, $v_2 approx 0$. (Light body bounces back with same speed, heavy body remains at rest)

* *Example:* A $0.1,\text{kg}$ ball at $5,\text{m/s}$ hits a stationary $100,\text{kg}$ block. The ball rebounds at $-5,\text{m/s}$, the block barely moves.

Massive Body with Light Body at Rest ($m_1 gg m_2$, $u_2 = 0$): — $v_1 approx u_1$, $v_2 approx 2u_1$. (Massive body continues, light body shoots forward at double speed)

* *Example:* A $100,\text{kg}$ block at $5,\text{m/s}$ hits a stationary $0.1,\text{kg}$ ball. The block continues at $5,\text{m/s}$, the ball moves at $10,\text{m/s}$.

Remember to use consistent sign conventions for velocities (e.g., right is positive, left is negative). Elastic collisions are idealizations, but their principles are vital for understanding energy and momentum transfer.

Prelims Revision Notes

Elastic collisions are defined by the conservation of both total linear momentum and total kinetic energy. This is a critical distinction from inelastic collisions where only momentum is conserved. The coefficient of restitution ($e$) for an elastic collision is always 1, indicating that the relative speed of separation after the collision is equal to the relative speed of approach before the collision ($|v_2 - v_1| = |u_1 - u_2|$). This relation is extremely useful for solving problems quickly.

For one-dimensional (head-on) elastic collisions, the general formulas for final velocities are: v_1 = left(\frac{m_1 - m_2}{m_1 + m_2}\right)u_1 + left(\frac{2m_2}{m_1 + m_2}\right)u_2 v_2 = left(\frac{2m_1}{m_1 + m_2}\right)u_1 + left(\frac{m_2 - m_1}{m_1 + m_2}\right)u_2

Key Special Cases to Memorize:

1
Equal Masses ($m_1 = m_2 = m$) and Second Body at Rest ($u_2 = 0$): — The first body comes to rest ($v_1 = 0$), and the second body moves with the initial velocity of the first ($v_2 = u_1$). This is a direct exchange of velocities.
2
Equal Masses ($m_1 = m_2 = m$) and Both Moving: — The bodies simply exchange their velocities ($v_1 = u_2$, $v_2 = u_1$).
3
Light Body Collides with Massive Body at Rest ($m_1 ll m_2$, $u_2 = 0$): — The light body reverses its direction with approximately the same speed ($v_1 approx -u_1$), and the massive body remains almost at rest ($v_2 approx 0$). Think of a ball bouncing off a wall.
4
Massive Body Collides with Light Body at Rest ($m_1 gg m_2$, $u_2 = 0$): — The massive body continues with nearly its original velocity ($v_1 approx u_1$), and the light body moves forward with approximately twice the initial velocity of the massive body ($v_2 approx 2u_1$).

Always use a consistent sign convention for velocities. If a problem asks for 'change in momentum' for a single body, remember it's $P_{\text{final}} - P_{\text{initial}}$. For a ball bouncing off a wall, the change in momentum is $-2mv$.

Vyyuha Quick Recall

For Elastic Collisions, remember 'MKE': Momentum is conserved. Kinetic energy is conserved. Equals 1 (Coefficient of Restitution, $e=1$).

And for the relative velocities: 'Approach = Separation' ($u_1 - u_2 = v_2 - v_1$).

For equal masses, 'Swap Speeds!'