Oblique Collisions in 2D: Elastic & Inelastic Collisions, Line of Impact & Glancing Collisions | JEE Physics Class 11

3. Total Post-Collision Speeds & Scattering Angles

• Net Final Speeds:
  $$v_1 = \sqrt{v_{1n}^2 + v_{1t}^2}, \qquad v_2 = \sqrt{v_{2n}^2 + v_{2t}^2}$$
• Deflection Angles relative to the Line of Impact:
  $$\tan\theta_1 = \frac{v_{1t}}{v_{1n}}, \qquad \tan\theta_2 = \frac{v_{2t}}{v_{2n}}$$

4. Classical JEE Theorem: Elastic Oblique Collision of Identical Masses ($m_1 = m_2 = m, e = 1$)

When a moving particle elastically strikes an identical stationary particle ($\vec{u}_2 = \vec{0}$):

• Linear momentum conservation: $\vec{u}_1 = \vec{v}_1 + \vec{v}_2 \implies u_1^2 = v_1^2 + v_2^2 + 2(\vec{v}_1 \cdot \vec{v}_2)$.
• Kinetic energy conservation: $\frac{1}{2}m u_1^2 = \frac{1}{2}m v_1^2 + \frac{1}{2}m v_2^2 \implies u_1^2 = v_1^2 + v_2^2$.
• Equating both relations yields:
  $$2(\vec{v}_1 \cdot \vec{v}_2) = 0 \implies \mathbf{\vec{v}_1 \cdot \vec{v}_2 = 0}$$
• Fundamental Result: The two identical particles emerge at right angles ($90^\circ$) to each other ($\theta_1 + \theta_2 = 90^\circ$), provided the collision is glancing (non-head-on, $v_1 \ne 0$)!
• Inelastic Extension: If $0 \le e < 1$, the angle between their trajectories is acute: $\theta_1 + \theta_2 < 90^\circ$.

5. Geometry of the Impact Parameter ($b$)

For two identical smooth spheres of radius $R$:

• The center-to-center distance at contact is $d = 2R$.
• If the perpendicular distance between their initial lines of motion is $b$ (the impact parameter, $0 \le b \le 2R$):
  $$\sin\alpha = \frac{b}{2R}, \qquad \cos\alpha = \sqrt{1 – \left(\frac{b}{2R}\right)^2}$$
  where $\alpha$ is the inclination of the line of centers to the incident line of motion.

6. Oblique Collision with a Fixed Rigid Smooth Wall

A particle strikes a smooth fixed wall with speed $u$ at angle of incidence $\alpha$ to the normal:

• Tangential component unchanged: $v_t = u_t = u \sin\alpha$.
• Normal component reversed and scaled by $e$: $v_n = e u_n = e u \cos\alpha$.
• Rebound angle $\beta$ with the normal:
  $$\mathbf{\tan\beta = \frac{v_t}{v_n} = \frac{u\sin\alpha}{e u\cos\alpha} = \frac{\tan\alpha}{e}}$$
• If $e < 1$, then $\tan\beta > \tan\alpha \implies \beta > \alpha$ (the rebound trajectory is flattened closer to the surface).
• Rebound speed: $v = \sqrt{v_n^2 + v_t^2} = u\sqrt{e^2 \cos^2\alpha + \sin^2\alpha}$.
• Impulse delivered to the wall: $J_n = m(1 + e)u\cos\alpha$.

7. Common JEE Pitfalls & Traps

• Trap 1 (Wrong Axis for Restitution): The coefficient of restitution $e$ applies ONLY to velocity components along the line of impact (common normal). Applying $e$ to total speeds or along the incident trajectory is invalid.
• Trap 2 (Misunderstanding Tangential Momentum): Total momentum is conserved in all directions, but individual momentum is conserved ONLY parallel to the common frictionless tangent.
• Trap 3 (Angle of Reflection Equality): $\beta = \alpha$ holds ONLY for perfectly elastic wall impacts ($e = 1$). If $e < 1$, the rebound angle with the normal is strictly greater than the incident angle ($\beta > \alpha$).

Solved Examples

Example 1 (Direct Application – Smooth Wall Impact):
A sphere of mass $m = 0.50\text{ kg}$ traveling at $u = 20.0\text{ m/s}$ strikes a smooth vertical wall at an angle of incidence $\alpha$ such that $\cos\alpha = 0.80$ and $\sin\alpha = 0.60$ with the normal. The coefficient of restitution is $e = 0.5625 = 9/16$.
(a) Find the normal and tangential components of the rebound velocity.
(b) Calculate the final speed $v$ and the angle of reflection $\beta$ with the normal.
(c) Determine the magnitude of the impulse delivered to the wall and the loss in kinetic energy.

Solution:
(a) Initial velocity components relative to the wall normal and tangent:
– Normal component toward wall: $u_n = u \cos\alpha = 20.0(0.80) = 16.0\text{ m/s}$.
– Tangential component parallel to wall: $u_t = u \sin\alpha = 20.0(0.60) = 12.0\text{ m/s}$.
Since the wall is smooth, there is no tangential impulse:
$$v_t = u_t = 12.0\text{ m/s}$$
The normal component rebounds according to the restitution coefficient:
$$v_n = e u_n = \left(\frac{9}{16}\right)(16.0) = 9.0\text{ m/s}$$

(b) Net rebound speed:
$$v = \sqrt{v_n^2 + v_t^2} = \sqrt{9.0^2 + 12.0^2} = \sqrt{81 + 144} = \sqrt{225} = 15.0\text{ m/s}$$
Angle of reflection with the normal:
$$\tan\beta = \frac{v_t}{v_n} = \frac{12.0}{9.0} = \frac{4}{3} \implies \beta = \arctan(4/3) \approx 53.13^\circ$$

(c) Normal impulse delivered to the wall:
$$J_n = m(v_n – (-u_n)) = m(v_n + u_n) = 0.50(9.0 + 16.0) = 0.50(25.0) = 12.5\text{ N}\cdot\text{s}$$
Kinetic energy loss:
$$K_i = \frac{1}{2}m u^2 = \frac{1}{2}(0.50)(20.0)^2 = 100.0\text{ J}$$
$$K_f = \frac{1}{2}m v^2 = \frac{1}{2}(0.50)(15.0)^2 = 56.25\text{ J}$$
$$\Delta K_{\text{loss}} = K_i – K_f = 100.0 – 56.25 = 43.75\text{ Joules}$$

Example 2 (Mathematical Derivation – Impact Parameter & Scattering):
A moving billiard ball $A$ of radius $R$ and mass $m$ strikes an identical stationary billiard ball $B$ with speed $u$. The collision is perfectly elastic ($e = 1$) and smooth, with an impact parameter $b = R$.
(a) Find the angle $\alpha$ made by the line of impact with the initial line of motion.
(b) Determine the final velocity vectors $\vec{v}_A$ and $\vec{v}_B$.
(c) Verify that the angle between their post-collision trajectories is $90^\circ$.

Solution:
(a) At the instant of contact, the distance between sphere centers is $d = 2R$.
From geometry:
$$\sin\alpha = \frac{b}{2R} = \frac{R}{2R} = \frac{1}{2} \implies \alpha = 30^\circ$$
$$\cos\alpha = \cos 30^\circ = \frac{\sqrt{3}}{2}$$

(b) Establish axes: Let $\hat{n}$ be the line of impact (directed from $A$’s center to $B$’s center at angle $-30^\circ$ to the initial velocity), and $\hat{t}$ be the common tangent ($+60^\circ$ to initial velocity).
Resolve initial velocity of ball $A$:
$$u_{An} = u\cos\alpha = u\cos 30^\circ = \frac{\sqrt{3}}{2}u$$
$$u_{At} = u\sin\alpha = u\sin 30^\circ = \frac{1}{2}u$$
For stationary ball $B$: $u_{Bn} = 0, u_{Bt} = 0$.
Since masses are identical ($m_A = m_B$) and $e = 1$, velocities along $\hat{n}$ are completely exchanged:
$$v_{An} = u_{Bn} = 0, \qquad v_{Bn} = u_{An} = \frac{\sqrt{3}}{2}u$$
Along the common tangent $\hat{t}$:
$$v_{At} = u_{At} = \frac{1}{2}u, \qquad v_{Bt} = u_{Bt} = 0$$
Hence:
– Ball $A$ moves purely along the common tangent with speed $v_A = \frac{1}{2}u$.
– Ball $B$ moves purely along the line of centers with speed $v_B = \frac{\sqrt{3}}{2}u$.

(c) Because ball $A$ moves along the common tangent $\hat{t}$ and ball $B$ moves along the normal $\hat{n}$, and the tangent is strictly perpendicular to the normal ($\hat{t} \perp \hat{n}$), the angle between their trajectories is identically $90^\circ$.
Check energy: $v_A^2 + v_B^2 = \left(\frac{u}{2}\right)^2 + \left(\frac{\sqrt{3}u}{2}\right)^2 = \frac{u^2}{4} + \frac{3u^2}{4} = u^2$. Kinetic energy is 100% conserved.

Example 3 (Standard JEE Advanced Scenario – Right-Angle Deflection Condition):
A particle of mass $m_1$ moving with speed $u_1$ along $+x$ strikes a stationary target particle of mass $m_2$ in an elastic oblique collision ($e = 1$). Particle $m_1$ is deflected through an angle $\theta_1 = 90^\circ$ relative to its initial direction of motion.
(a) Prove that such a right-angle deflection is dynamically possible only if $m_1 < m_2$.
(b) Derive the final speeds $v_1$ and $v_2$ in terms of $u_1, m_1,$ and $m_2$.
(c) Find the angle $\theta_2$ at which the target particle $m_2$ recoils.

Solution:
(a) Choose the initial motion along $+x$. After collision, particle $1$ moves along $+y$, so $\vec{v}_1 = v_1\hat{j}$.
By linear momentum conservation:
– $x$-component: $m_1 u_1 = m_2 v_{2x} \implies v_{2x} = \frac{m_1}{m_2}u_1$.
– $y$-component: $0 = m_1 v_1 + m_2 v_{2y} \implies v_{2y} = -\frac{m_1}{m_2}v_1$.
The total speed of $m_2$ satisfies:
$$v_2^2 = v_{2x}^2 + v_{2y}^2 = \frac{m_1^2}{m_2^2}(u_1^2 + v_1^2)$$
By conservation of kinetic energy ($e = 1$):
$$\frac{1}{2}m_1 u_1^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \implies m_2 v_2^2 = m_1(u_1^2 – v_1^2)$$
Substituting $v_2^2$:
$$m_2 \left[\frac{m_1^2}{m_2^2}(u_1^2 + v_1^2)\right] = m_1(u_1^2 – v_1^2) \implies \frac{m_1}{m_2}(u_1^2 + v_1^2) = u_1^2 – v_1^2$$
Letting mass ratio $r = \frac{m_1}{m_2}$:
$$r u_1^2 + r v_1^2 = u_1^2 – v_1^2 \implies (1 + r)v_1^2 = (1 – r)u_1^2 \implies v_1^2 = \left(\frac{1 – r}{1 + r}\right)u_1^2 = \left(\frac{m_2 – m_1}{m_2 + m_1}\right)u_1^2$$
For a real, non-zero speed $v_1$, the numerator must be strictly positive:
$$m_2 – m_1 > 0 \implies \mathbf{m_1 < m_2}$$

(b) Final speeds:
$$v_1 = u_1\sqrt{\frac{m_2 – m_1}{m_2 + m_1}}$$
Substituting $v_1^2$ into the kinetic energy equation:
$$v_2^2 = \frac{m_1}{m_2}(u_1^2 – v_1^2) = \frac{m_1}{m_2}u_1^2\left[1 – \frac{m_2 – m_1}{m_2 + m_1}\right] = \frac{m_1}{m_2}u_1^2\left[\frac{2m_1}{m_2 + m_1}\right] \implies v_2 = u_1\sqrt{\frac{2m_1^2}{m_2(m_1 + m_2)}}$$

(c) Recoil direction of target $m_2$:
$$\tan\theta_2 = \frac{|v_{2y}|}{v_{2x}} = \frac{\frac{m_1}{m_2}v_1}{\frac{m_1}{m_2}u_1} = \frac{v_1}{u_1} = \sqrt{\frac{m_2 – m_1}{m_2 + m_1}}$$

Example 4 (Edge Case – Inelastic Oblique Impact of Two Identical Spheres):
Two identical smooth spheres $1$ and $2$ of mass $m$ each lie on a horizontal plane. Sphere $1$ moves with velocity $\vec{u}_1 = (6.0\hat{i})\text{ m/s}$ and strikes stationary sphere $2$. At the instant of contact, the unit normal pointing from sphere $1$ to sphere $2$ is $\hat{n} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$. The coefficient of restitution is $e = 0.60$.
(a) Resolve the initial velocity into normal and tangential components.
(b) Calculate the post-collision velocity vectors $\vec{v}_1$ and $\vec{v}_2$.
(c) Determine the total kinetic energy loss $\Delta K_{\text{loss}}$ for $m = 1.0\text{ kg}$.

Solution:
(a) Unit vectors:
$$\hat{n} = \frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}, \qquad \hat{t} = -\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}$$
Resolving initial velocity $\vec{u}_1 = 6.0\hat{i}\text{ m/s}$:
$$u_{1n} = \vec{u}_1 \cdot \hat{n} = 6.0\left(\frac{\sqrt{3}}{2}\right) = 3\sqrt{3}\text{ m/s}$$
$$u_{1t} = \vec{u}_1 \cdot \hat{t} = 6.0\left(-\frac{1}{2}\right) = -3.0\text{ m/s}$$
For sphere $2$ initially at rest: $u_{2n} = 0, u_{2t} = 0$.

(b) Along the normal direction $\hat{n}$ ($m_1 = m_2 = m, e = 0.60, u_{2n} = 0$):
$$v_{1n} = \left(\frac{1 – e}{2}\right)u_{1n} = \left(\frac{1 – 0.60}{2}\right)(3\sqrt{3}) = 0.20(3\sqrt{3}) = 0.6\sqrt{3}\text{ m/s}$$
$$v_{2n} = \left(\frac{1 + e}{2}\right)u_{1n} = \left(\frac{1 + 0.60}{2}\right)(3\sqrt{3}) = 0.80(3\sqrt{3}) = 2.4\sqrt{3}\text{ m/s}$$
Along the frictionless tangent direction $\hat{t}$:
$$v_{1t} = u_{1t} = -3.0\text{ m/s}, \qquad v_{2t} = u_{2t} = 0$$
Reconstructing Cartesian velocity vectors:
$$\vec{v}_1 = v_{1n}\hat{n} + v_{1t}\hat{t} = (0.6\sqrt{3})\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) + (-3.0)\left(-\frac{1}{2}\hat{i} + \frac{\sqrt{3}}{2}\hat{j}\right) = 2.4\hat{i} – 1.2\sqrt{3}\hat{j}\text{ m/s}$$
$$\vec{v}_2 = v_{2n}\hat{n} + v_{2t}\hat{t} = (2.4\sqrt{3})\left(\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) + \vec{0} = 3.6\hat{i} + 1.2\sqrt{3}\hat{j}\text{ m/s}$$
Check total momentum: $\vec{v}_1 + \vec{v}_2 = (2.4 + 3.6)\hat{i} + (-1.2\sqrt{3} + 1.2\sqrt{3})\hat{j} = 6.0\hat{i} = \vec{u}_1$.

(c) Kinetic energy loss occurs purely along the normal line of impact:
$$\Delta K_{\text{loss}} = \frac{1}{2}\mu (u_{1n} – u_{2n})^2 (1 – e^2)$$
Here $\mu = \frac{m \times m}{m + m} = \frac{m}{2} = 0.50\text{ kg}$.
$$\Delta K_{\text{loss}} = \frac{1}{2}(0.50)(3\sqrt{3})^2 [1 – (0.60)^2] = \frac{1}{4}(27)(1 – 0.36) = 4.32\text{ Joules}$$

Worksheet: 10 Practice Problems

Problem 1 (JEE Main – Single Correct):
Two identical smooth balls $A$ and $B$ undergo a perfectly elastic 2D collision with ball $B$ initially at rest. If the collision is glancing (non-head-on), the angle between their post-collision velocity vectors is:
(A) $45^\circ$
(B) $60^\circ$
(C) $90^\circ$
(D) $120^\circ$

Problem 2 (JEE Main – Single Correct):
A small ball strikes a smooth fixed vertical wall with speed $u$ at an angle of incidence $\theta$ to the normal. If the coefficient of restitution is $e$, the angle of rebound $\phi$ with the normal satisfies:
(A) $\tan\phi = e\tan\theta$
(B) $\tan\phi = \frac{\tan\theta}{e}$
(C) $\tan\phi = \tan\theta$
(D) $\sin\phi = e\sin\theta$

Problem 3 (JEE Main – Single Correct):
In an oblique collision between two smooth bodies, the component of velocity of each body that remains individually conserved is:
(A) Along the line of impact (common normal)
(B) Along the common tangent
(C) Perpendicular to the plane of collision
(D) In the direction of initial relative velocity

Problem 4 (JEE Main – Single Correct):
A ball strikes a smooth floor with speed $u$ at an angle of $45^\circ$ to the vertical. If the coefficient of restitution is $e = \frac{1}{\sqrt{3}}$, the speed of the ball immediately after reflection is:
(A) $\frac{u}{\sqrt{2}}$
(B) $u\sqrt{\frac{2}{3}}$
(C) $\frac{u}{2}$
(D) $\frac{u}{\sqrt{3}}$

Problem 5 (JEE Main – Single Correct):
Two identical smooth spheres each of radius $R$ collide elastically. Sphere 1 has initial speed $u$ and sphere 2 is stationary. If the impact parameter is $b = \sqrt{3}R$, the angle $\alpha$ made by the line of centers with the initial velocity is:
(A) $30^\circ$
(B) $45^\circ$
(C) $60^\circ$
(D) $90^\circ$

Problem 6 (JEE Advanced – One or More Correct):
A ball of mass $m$ strikes a smooth horizontal floor with velocity $\vec{u} = u_x\hat{i} – u_y\hat{j}$ ($u_y > 0$). The coefficient of restitution is $e$ ($0 < e < 1$). Which of the following statements is/are correct?
(A) The velocity vector immediately after rebound is $\vec{v} = u_x\hat{i} + e u_y\hat{j}$.
(B) The impulse delivered to the floor is $\vec{J} = m(1 + e)u_y\hat{j}$.
(C) The kinetic energy dissipated during the collision is $\Delta K = \frac{1}{2}m u_y^2(1 – e^2)$.
(D) The angle made by the rebound velocity with the horizontal floor is less than the angle made by the incident velocity with the horizontal floor.

Problem 7 (JEE Advanced – One or More Correct):
Two smooth spheres of masses $m_1$ and $m_2$ undergo an oblique collision on a frictionless horizontal table. Which of the following quantities is/are strictly conserved during the impact?
(A) Total linear momentum of the two-sphere system along the line of impact
(B) Total linear momentum of the two-sphere system along the common tangent
(C) Linear momentum of sphere 1 along the common tangent
(D) Total kinetic energy associated with motion along the common tangent

Problem 8 (JEE Advanced – One or More Correct):
In an oblique collision between two identical smooth particles ($m_1 = m_2$) where the target is initially stationary ($u_2 = 0$), with coefficient of restitution $e$:
(A) If $e = 1$, the angle between their final velocity vectors is always $90^\circ$ (unless head-on).
(B) If $e < 1$, the angle between their final velocity vectors is strictly acute ($< 90^\circ$).
(C) The target particle always recoils along the line of impact immediately after collision.
(D) If $e = 0$, both particles move in identical directions after collision.

Problem 9 (JEE Main / Advanced – Numerical Value Type):
A ball strikes a smooth fixed floor with speed $u = 20.0\text{ m/s}$ at an angle of incidence $\alpha$ with the normal such that $\cos\alpha = 0.80$ and $\sin\alpha = 0.60$. If the coefficient of restitution is $e = \frac{9}{16} = 0.5625$, calculate the speed of the ball after reflection in $\text{m/s}$.

Problem 10 (JEE Main / Advanced – Numerical Value Type):
Two identical smooth balls $A$ and $B$ undergo an elastic oblique collision ($e = 1$). Ball $A$ moves with initial speed $u = 10.0\text{ m/s}$ and strikes stationary ball $B$. If ball $A$ is deflected through an angle of $60^\circ$ relative to its original direction of motion, calculate the final speed of ball $A$ in $\text{m/s}$.

Solutions & Explanations

Answer Key Summary:
1. (C) | 2. (B) | 3. (B) | 4. (B) | 5. (C) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, C) | 9. 15 | 10. 5

Solution 1:
For two identical particles in an elastic collision with one at rest:
$\vec{u}_A = \vec{v}_A + \vec{v}_B \implies u_A^2 = v_A^2 + v_B^2 + 2\vec{v}_A \cdot \vec{v}_B$.
By kinetic energy conservation: $u_A^2 = v_A^2 + v_B^2$.
Equating gives $\vec{v}_A \cdot \vec{v}_B = 0 \implies \vec{v}_A \perp \vec{v}_B$, so the angle between their trajectories is $90^\circ$.
Correct Option: (C)

Solution 2:
Tangential component: $v_t = u_t = u\sin\theta$.
Normal component: $v_n = e u_n = e u\cos\theta$.
The rebound angle with the normal is given by:
$$\tan\phi = \frac{v_t}{v_n} = \frac{u\sin\theta}{e u\cos\theta} = \frac{\tan\theta}{e}$$
Correct Option: (B)

Solution 3:
Because the surfaces are smooth and frictionless, mutual contact forces act purely along the line of impact (normal). Zero impulsive force acts parallel to the common tangent, so individual tangential momentum and velocity are conserved for each body: $v_{1t} = u_{1t}$ and $v_{2t} = u_{2t}$.
Correct Option: (B)

Solution 4:
Incident angle with normal (vertical) is $\alpha = 45^\circ$.
$u_t = u\sin 45^\circ = \frac{u}{\sqrt{2}}$.
$u_n = u\cos 45^\circ = \frac{u}{\sqrt{2}}$.
$v_t = u_t = \frac{u}{\sqrt{2}}$.
$v_n = e u_n = \left(\frac{1}{\sqrt{3}}\right)\frac{u}{\sqrt{2}} = \frac{u}{\sqrt{6}}$.
Rebound speed:
$$v = \sqrt{v_t^2 + v_n^2} = \sqrt{\frac{u^2}{2} + \frac{u^2}{6}} = \sqrt{\frac{4u^2}{6}} = u\sqrt{\frac{2}{3}}$$
Correct Option: (B)

Solution 5:
At the instant of impact, center-to-center distance is $d = 2R$.
$$\sin\alpha = \frac{b}{2R} = \frac{\sqrt{3}R}{2R} = \frac{\sqrt{3}}{2} \implies \alpha = 60^\circ$$
Correct Option: (C)

Solution 6:
– (A) True: Floor exerts force along $+\hat{j}$; tangential velocity along $\hat{i}$ is unaffected, while vertical component rebounds as $e u_y\hat{j}$.
– (B) True: Impulse on floor: $\vec{J} = -\Delta \vec{p}_{\text{ball}} = -m(e u_y\hat{j} – (-u_y\hat{j})) = m(1 + e)u_y\hat{j}$.
– (C) True: Dissipation occurs only in normal motion: $\Delta K = \frac{1}{2}m u_y^2 – \frac{1}{2}m(e u_y)^2 = \frac{1}{2}m u_y^2(1 – e^2)$.
– (D) True: Angle with horizontal $\theta_h$ satisfies $\tan\theta_h = \frac{v_y}{v_x} = \frac{e u_y}{u_x} < \frac{u_y}{u_x}$, so the rebound angle with the floor is flattened.
Correct Options: (A, B, C, D)

Solution 7:
– (A) True: No external force acts along $\hat{n}$, so total normal momentum is conserved.
– (B) True: Total tangential momentum is conserved.
– (C) True: Frictionless contact means zero tangential force on sphere 1 individually, conserving $m_1 v_{1t}$.
– (D) True: Since $v_{1t} = u_{1t}$ and $v_{2t} = u_{2t}$, tangential kinetic energy $\frac{1}{2}m_1 v_{1t}^2 + \frac{1}{2}m_2 v_{2t}^2$ is identical before and after impact.
Correct Options: (A, B, C, D)

Solution 8:
– (A) True: $\vec{v}_1 \cdot \vec{v}_2 = 0 \implies 90^\circ$ divergence for elastic collisions.
– (B) True: $\vec{v}_1 \cdot \vec{v}_2 = v_{1n} v_{2n} + v_{1t} v_{2t} = \left(\frac{1-e}{2}u_n\right)\left(\frac{1+e}{2}u_n\right) + 0 = \frac{1-e^2}{4}u_n^2 > 0$ for $e < 1$, proving the dot product is positive and the angle is acute ($< 90^\circ$).
– (C) True: Target is initially at rest ($u_{2t} = 0$) and receives impulse only along $\hat{n}$, so $C) True: Target is initially at rest ($u_{2t} = 0$) and receives impulse only along $\hat{n}$, so $v_{2t} = 0$; it moves purely along the line of impact.
– (D) False: When $e = 0$, $v_{1n} = v_{2n}$, but $v_{1t} = u_{1t} \ne 0$ while $v_{2t} = 0$, so their velocity vectors point in different directions.
Correct Options: (A, B, C)

Solution 9:
$u_t = u\sin\alpha = 20.0 \times 0.60 = 12.0\text{ m/s}$.
$u_n = u\cos\alpha = 20.0 \times 0.80 = 16.0\text{ m/s}$.
$v_t = u_t = 12.0\text{ m/s}$.
$v_n = e u_n = \left(\frac{9}{16}\right)(16.0) = 9.0\text{ m/s}$.
Rebound speed:
$$v = \sqrt{v_t^2 + v_n^2} = \sqrt{12.0^2 + 9.0^2} = \sqrt{144 + 81} = \sqrt{225} = 15.0\text{ m/s}$$
Correct Answer: 15

Solution 10:
In an elastic oblique collision between equal masses with the target at rest, the two particles separate at $90^\circ$:
$\vec{u} = \vec{v}_A + \vec{v}_B$ with $\vec{v}_A \perp \vec{v}_B$.
This forms a right triangle where $\vec{u}$ is the hypotenuse:
$$v_A = u \cos\theta_A$$
Given $u = 10.0\text{ m/s}$ and deflection angle $\theta_A = 60^\circ$:
$$v_A = 10.0 \cos 60^\circ = 10.0 \times 0.50 = 5.0\text{ m/s}$$
Correct Answer: 5