Tag: 2-D collisions

Summary

This is Lecture 12 (last lecture) 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 Elastic collision and inelastic collisions in Two Dimensions and different aspects of it.

1. Introduction to Oblique Collisions

When two bodies collide such that their initial velocities are not along the line joining their centers of mass at the instant of impact, the collision is called an oblique collision (or 2D collision).

Unlike head-on (1D) collisions where motion remains confined to a single straight line, oblique collisions cause the bodies to scatter at various angles in a plane. However, the fundamental laws of physics—Conservation of Linear Momentum and the Coefficient of Restitution—still govern the interaction, provided we analyze them along the correct coordinate axes.

2. The Coordinate Frame: Line of Impact vs. Plane of Contact

To solve 2D collision problems efficiently, we resolve all velocity vectors into two mutually perpendicular directions relative to the colliding surfaces:

1. The Line of Impact (LOI): The common normal line passing through the centers of mass of the two bodies at the exact instant of impact.

• Mutual impulsive forces (and normal reaction) act only along this line.
• Linear momentum is conserved along the LOI.
• The Coefficient of Restitution ($e$) applies strictly along this line:
  $$v_{2n} – v_{1n} = e(u_{1n} – u_{2n})$$
  (where subscript $n$ denotes components along the Line of Impact).

2. The Plane of Contact (POC) / Tangential Direction: The plane perpendicular to the Line of Impact at the point of contact.

• Assuming the colliding surfaces are smooth and frictionless, no tangential forces or impulses act between the bodies during impact.
• Therefore, the component of velocity for each body along the tangential direction remains completely unchanged before and after the collision:
  $$v_{1t} = u_{1t} \quad \text{and} \quad v_{2t} = u_{2t}$$

3. Step-by-Step Problem Solving Strategy for JEE-Mains

When tackling an oblique collision problem, execute these steps systematically:

• Step 1: Identify the Line of Impact (LOI) by drawing the line connecting the centers of mass at impact (or taking the normal vector given in the problem).
• Step 2: Resolve the initial velocity vectors $\vec{u}_1$ and $\vec{u}_2$ of both bodies into two components:

• Normal component along the LOI ($u_{1n}, u_{2n}$).
• Tangential component along the POC ($u_{1t}, u_{2t}$).

• Step 3: Set the tangential final velocities equal to the initial ones: $v_{1t} = u_{1t}$ and $v_{2t} = u_{2t}$.
• Step 4: Apply Conservation of Linear Momentum along the LOI:
  $$m_1 u_{1n} + m_2 u_{2n} = m_1 v_{1n} + m_2 v_{2n}$$
• Step 5: Apply the Restitution Equation along the LOI:
  $$v_{2n} – v_{1n} = e(u_{1n} – u_{2n})$$
• Step 6: Solve the simultaneous equations for $v_{1n}$ and $v_{2n}$, then reconstruct the final velocity vectors: $\vec{v}_1 = v_{1n}\hat{n} + v_{1t}\hat{t}$ and $\vec{v}_2 = v_{2n}\hat{n} + v_{2t}\hat{t}$.

4. High-Yield Special Case: Perfectly Elastic Collision of Equal Masses

Consider two spheres of equal mass ($m_1 = m_2 = m$) undergoing a perfectly elastic ($e=1$) oblique collision, where one sphere is initially at rest ($u_2 = 0$).

• Applying momentum conservation and restitution along the LOI for equal masses yields an exchange of normal components: $v_{1n} = u_{2n} = 0$ and $v_{2n} = u_{1n}$.
• Because tangential components are unchanged, the first sphere stops moving along the LOI and continues purely along its original tangential line, while the second sphere moves off along the LOI.
• The Golden Rule: When two equal masses undergo a perfectly elastic oblique collision and one is initially stationary, the angle between their final velocity vectors is always $90^\circ$ ($\pi/2$ radians).

JEE-Mains Practice Questions

Question 1: Oblique Collision with a Fixed Wall

A smooth ball of mass $m$ strikes a rigid, fixed vertical wall at an angle of incidence $\theta = 30^\circ$ to the normal, with a speed $u$. If the coefficient of restitution between the ball and the wall is $e = 0.5$, find the angle of rebound $\beta$ with the normal.

Solution:

1. Choose the Line of Impact (LOI) perpendicular to the wall (normal direction) and the Plane of Contact (POC) parallel to the wall (tangential direction).
2. Initial normal component: $u_n = u \cos(30^\circ)$. Initial tangential component: $u_t = u \sin(30^\circ)$.
3. Tangential velocity remains unchanged: $v_t = u_t = u \sin(30^\circ)$.
4. Normal velocity after rebound using $e$:
   $v_n = e \cdot u_n = 0.5 \times u \cos(30^\circ) = \frac{1}{2} u \left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4}u$.
5. Find the rebound angle $\beta$:
   $\tan\beta = \frac{v_t}{v_n} = \frac{u \sin(30^\circ)}{\frac{\sqrt{3}}{4}u} = \frac{0.5}{\frac{\sqrt{3}}{4}} = \frac{2}{\sqrt{3}}$.
   $\beta = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right)$.

Question 2: Two-Particle Oblique Elastic Collision

A particle of mass $m$ moving with velocity $u$ along the x-axis strikes another identical stationary particle. After the elastic collision, one particle moves at an angle of $30^\circ$ to the x-axis. What is the angle of motion of the second particle?

Solution:

1. This is a perfectly elastic collision between equal masses ($m_1 = m_2 = m$) where one body is initially at rest ($u_2 = 0$).
2. As established in the special properties of 2D elastic collisions for equal masses, the angle between the final velocity vectors of the two particles is always $90^\circ$.
3. If the first particle moves at $\theta_1 = 30^\circ$, the second particle must move at $\theta_2 = 90^\circ – 30^\circ = 60^\circ$ on the opposite side of the x-axis.
   Answer: $60^\circ$

Question 3: Component-Based Restitution

Two smooth spheres of equal mass $m$ collide obliquely. Just before impact, sphere A has a velocity vector $\vec{u}_A = (4\hat{i} + 3\hat{j}) \text{ m/s}$ and sphere B has $\vec{u}_B = (1\hat{i} – 1\hat{j}) \text{ m/s}$. At the moment of impact, the Line of Impact is parallel to the x-axis ($\hat{i}$). If the coefficient of restitution is $e = 0.5$, find the final velocity vector of sphere A.

Solution:

1. The LOI is along the x-axis ($\hat{i}$), so x-components are normal ($n$) and y-components are tangential ($t$).
2. Initial components for A: $u_{An} = 4$, $u_{At} = 3$.
3. Initial components for B: $u_{Bn} = 1$, $u_{Bt} = -1$.
4. Tangential velocities remain unchanged:
   $v_{At} = u_{At} = 3$ (so $y$-component of $\vec{v}_A$ is $3\hat{j}$).
5. Apply momentum conservation and restitution along the LOI (x-axis) for equal masses ($m_1 = m_2 = m$):
   Since masses are equal, the normal velocities simply update via the standard elastic/inelastic exchange formula:
   $v_{An} = \frac{(m – em)u_{An} + (1+e)m u_{Bn}}{2m} = \frac{(1 – e)u_{An} + (1+e)u_{Bn}}{2}$
   $v_{An} = \frac{(1 – 0.5)(4) + (1 + 0.5)(1)}{2} = \frac{0.5(4) + 1.5(1)}{2} = \frac{2 + 1.5}{2} = \frac{3.5}{2} = 1.75 \text{ m/s}$.
6. Final velocity vector for sphere A:
   $\vec{v}_A = 1.75\hat{i} + 3\hat{j} \text{ m/s}$.