Concept Card: Inelastic Collisions in One Dimension & Coefficient of Restitution
1. Physical Framework & Definitions:
An inelastic collision is an impact event in which the total linear momentum of the isolated system is strictly conserved, but the total macroscopic mechanical kinetic energy is NOT conserved ($K_f < K_i$). The lost kinetic energy is converted into microscopic internal thermal energy (heat), sound waves, or permanent plastic deformation work.
- Conservation of Linear Momentum: Because internal contact forces cancel out in action-reaction pairs (Newton’s Third Law), the net external force on the system during the brief collision duration is zero ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$):
$$\vec{p}_i = \vec{p}_f \iff m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2$$ - Newton’s Law of Restitution & Coefficient of Restitution ($e$):
For any two colliding bodies, the ratio of the relative speed of separation after collision to the relative speed of approach before collision, measured along the common normal (line of impact), is a scalar constant denoted by $e$:
$$e = \frac{\text{Relative Velocity of Separation}}{\text{Relative Velocity of Approach}} = \frac{v_2 – v_1}{u_1 – u_2}$$
(where $u_1 > u_2$ so that the bodies approach, and $v_2 > v_1$ so that they separate). - Classification of Collisions based on $e$:
- $e = 1$ (Perfectlys Elastic): Relative speed of separation equals relative speed of approach ($v_2 – v_1 = u_1 – u_2$). Mechanical kinetic energy is 100% conserved ($\Delta K = 0$).
- $0 < e < 1$ (Inelastic / Real-world): Separation speed is less than approach speed ($v_2 – v_1 < u_1 – u_2$). Some mechanical kinetic energy is dissipated into heat/deformation ($0 < \Delta K_{\text{loss}} < \Delta K_{\text{max}}$).
- $e = 0$ (Perfectlys Inelastic / Plastic): The bodies do not separate after impact; they stick together and move with a single common velocity ($v_1 = v_2 = v_{\text{cm}}$). The loss of kinetic energy is the maximum possible consistent with linear momentum conservation.
- $e > 1$ (Super-Elastic / Explosive): Internal potential or chemical energy is released during contact, resulting in a net increase in kinetic energy ($K_f > K_i$).
2. Rigorous Derivation of 1D Post-Collision Velocities
Consider two bodies of masses $m_1$ and $m_2$ moving along the $+x$-axis with initial velocities $u_1$ and $u_2$.
- Linear Momentum Conservation:
$$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \implies m_1 v_1 + m_2 v_2 = (m_1 + m_2)v_{\text{cm}}$$ - Restitution Equation:
$$v_2 – v_1 = e(u_1 – u_2) \implies v_2 = v_1 + e(u_1 – u_2)$$ - Master Velocity Formulas:
Substituting $v_2$ into the momentum equation yields:
$$\mathbf{v_1 = \left(\frac{m_1 – e m_2}{m_1 + m_2}\right)u_1 + \left(\frac{(1 + e)m_2}{m_1 + m_2}\right)u_2}$$
$$\mathbf{v_2 = \left(\frac{(1 + e)m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 – e m_1}{m_1 + m_2}\right)u_2}$$
3. Center of Mass (CoM) Frame Formulation (JEE Advanced Method)
In the Center of Mass frame, the velocity of the center of mass is $v_{\text{cm}} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$.
- Relative velocities in the CoM frame before impact:
$$u_{1,\text{cm}} = u_1 – v_{\text{cm}}, \qquad u_{2,\text{cm}} = u_2 – v_{\text{cm}}$$ - After collision, velocities in the CoM frame simply reverse direction and are scaled down by factor $e$:
$$v_{1,\text{cm}} = -e u_{1,\text{cm}} \implies \mathbf{v_1 = v_{\text{cm}} – e(u_1 – v_{\text{cm}})}$$
$$v_{2,\text{cm}} = -e u_{2,\text{cm}} \implies \mathbf{v_2 = v_{\text{cm}} – e(u_2 – v_{\text{cm}})}$$
4. General Loss of Kinetic Energy ($\Delta K_{\text{loss}}$)
The deficit between initial and final kinetic energies is:
$$\Delta K_{\text{loss}} = K_i – K_f = \frac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right)(u_1 – u_2)^2 (1 – e^2) = \frac{1}{2}\mu (u_{\text{rel}})^2 (1 – e^2)$$
where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass of the two-body system, and $u_{\text{rel}} = |u_1 – u_2|$ is the relative approach speed.
- If $e = 1 \implies \Delta K_{\text{loss}} = 0$ (Elastic collision).
- If $e = 0 \implies \Delta K_{\text{loss, max}} = \frac{1}{2}\mu (u_1 – u_2)^2$ (Perfectly inelastic collision).
5. Successive Bouncing of a Ball on a Fixed Floor
A ball dropped from initial height $h_0$ onto a stationary horizontal floor with restitution coefficient $e$:
- Speed just before 1st impact: $v_0 = \sqrt{2g h_0}$.
- Speed immediately after 1st bounce: $v_1 = e v_0$.
- Height attained after 1st bounce: $h_1 = \frac{v_1^2}{2g} = e^2 h_0$.
- Speed after $n$-th bounce: $v_n = e^n v_0 = e^n \sqrt{2g h_0}$.
- Height attained after $n$-th bounce: $h_n = e^{2n} h_0$.
- Total distance traveled before coming to rest:
$$D = h_0 + 2h_1 + 2h_2 + 2h_3 + \dots = h_0\left(\frac{1 + e^2}{1 – e^2}\right)$$ - Total time elapsed before coming to rest:
$$T = t_0 + 2t_1 + 2t_2 + 2t_3 + \dots = \sqrt{\frac{2h_0}{g}}\left(\frac{1 + e}{1 – e}\right)$$ - Average force exerted on the floor over the entire motion: $F_{\text{avg}} = \frac{\Delta p_{\text{total}}}{T} = mg$.
6. Common JEE Pitfalls & Traps
- Trap 1 (Kinetic Energy Loss in Perfectly Inelastic Collisions): Students often mistakenly assume that $100\%$ of kinetic energy is lost when $e = 0$. In an isolated system without external forces, the kinetic energy associated with center-of-mass motion $K_{\text{cm}} = \frac{1}{2}(m_1 + m_2)v_{\text{cm}}^2$ can NEVER be dissipated. Only internal relative kinetic energy $\frac{1}{2}\mu u_{\text{rel}}^2$ is lost.
- Trap 2 (Sign Conventions with $e$): In $e = \frac{v_2 – v_1}{u_1 – u_2}$, all velocities must be algebraic values along the same chosen coordinate axis. If a body rebounds in the reverse direction, its velocity must be entered with a negative sign.
- Trap 3 (Impulsive Normal vs. Non-Impulsive Gravity): During collision impact $\Delta t \to 0$, normal contact forces are enormous (impulsive), while continuous external forces like gravity $mg$ are negligible during the collision itself, justifying linear momentum conservation during the impact.
Solved Examples
Example 1 (Direct Application – Velocity & Energy Dissipation):
A body of mass $m_1 = 3.0\text{ kg}$ moving with velocity $u_1 = 8.0\text{ m/s}$ along $+x$ undergoes a head-on collision with a stationary body of mass $m_2 = 1.0\text{ kg}$. The coefficient of restitution is $e = 0.50$.
(a) Determine the velocities $v_1$ and $v_2$ of both bodies after the collision.
(b) Calculate the total initial kinetic energy, total final kinetic energy, and the mechanical energy dissipated.
(c) Verify the energy dissipation using the reduced mass formula $\Delta K = \frac{1}{2}\mu u_{\text{rel}}^2 (1 – e^2)$.
Solution:
(a) Given $m_1 = 3.0\text{ kg}, u_1 = +8.0\text{ m/s}, m_2 = 1.0\text{ kg}, u_2 = 0, e = 0.50$.
Applying the master velocity equations:
$$v_1 = \left(\frac{m_1 – e m_2}{m_1 + m_2}\right)u_1 = \left(\frac{3.0 – (0.50)(1.0)}{3.0 + 1.0}\right)(8.0) = \left(\frac{2.5}{4.0}\right)(8.0) = +5.0\text{ m/s}$$
$$v_2 = \left(\frac{(1 + e)m_1}{m_1 + m_2}\right)u_1 = \left(\frac{(1 + 0.50)(3.0)}{3.0 + 1.0}\right)(8.0) = \left(\frac{4.5}{4.0}\right)(8.0) = +9.0\text{ m/s}$$
Both bodies continue moving in the $+x$-direction, with $v_1 = 5.0\text{ m/s}$ and $v_2 = 9.0\text{ m/s}$.
Check restitution: $e = \frac{v_2 – v_1}{u_1 – u_2} = \frac{9.0 – 5.0}{8.0 – 0} = \frac{4.0}{8.0} = 0.50$ (Exact match!).
(b) Initial kinetic energy:
$$K_i = \frac{1}{2}m_1 u_1^2 + 0 = \frac{1}{2}(3.0)(8.0)^2 = 1.5 \times 64 = 96.0\text{ J}$$
Final kinetic energy:
$$K_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(3.0)(5.0)^2 + \frac{1}{2}(1.0)(9.0)^2 = 1.5(25) + 0.5(81) = 37.5 + 40.5 = 78.0\text{ J}$$
Mechanical energy dissipated:
$$\Delta K_{\text{loss}} = K_i – K_f = 96.0 – 78.0 = 18.0\text{ Joules}$$
(c) Using reduced mass:
$$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{3.0 \times 1.0}{4.0} = 0.75\text{ kg}$$
$$\Delta K_{\text{loss}} = \frac{1}{2}\mu (u_1 – u_2)^2 (1 – e^2) = \frac{1}{2}(0.75)(8.0)^2 [1 – (0.50)^2] = \frac{1}{2}(0.75)(64)(1 – 0.25) = 24 \times 0.75 = 18.0\text{ Joules}$$
Both methods produce identical results.
Example 2 (Mathematical Manipulation – Infinite Bouncing Series):
A rubber ball is released from rest from a height $h_0 = 10.0\text{ m}$ above a horizontal floor. The coefficient of restitution between the ball and floor is $e = 0.60$. Taking $g = 10.0\text{ m/s}^2$:
(a) Find the maximum height reached after the second bounce.
(b) Calculate the total cumulative distance traveled by the ball before it comes to rest.
(c) Calculate the total time elapsed before all bouncing ceases.
Solution:
(a) Rebound height after $n$ bounces: $h_n = e^{2n} h_0$.
For the 2nd bounce ($n = 2$):
$$h_2 = e^4 h_0 = (0.60)^4 (10.0) = 0.1296 \times 10.0 = 1.296\text{ meters}$$
(b) Total distance traveled:
$$D = h_0\left(\frac{1 + e^2}{1 – e^2}\right) = 10.0\left(\frac{1 + 0.60^2}{1 – 0.60^2}\right) = 10.0\left(\frac{1 + 0.36}{1 – 0.36}\right) = 10.0\left(\frac{1.36}{0.64}\right) = 10.0 \times 2.125 = 21.25\text{ meters}$$
(c) Total time taken:
Time for initial fall: $t_0 = \sqrt{\frac{2h_0}{g}} = \sqrt{\frac{2(10.0)}{10.0}} = \sqrt{2} \approx 1.414\text{ s}$.
Total time:
$$T = t_0\left(\frac{1 + e}{1 – e}\right) = \sqrt{2}\left(\frac{1 + 0.60}{1 – 0.60}\right) = \sqrt{2}\left(\frac{1.60}{0.40}\right) = 4\sqrt{2} \approx 5.657\text{ seconds}$$
Example 3 (Standard JEE Advanced Scenario – The Ballistic Pendulum):
A rifle bullet of mass $m = 20.0\text{ g} = 0.020\text{ kg}$ is fired horizontally with muzzle velocity $u$ into a wooden block of mass $M = 1.98\text{ kg}$ suspended by a light cord of length $L = 2.0\text{ m}$. The bullet embeds completely inside the block ($e = 0$). The combined mass then swings upward, reaching a maximum angular deflection $\theta = 60^\circ$. (Take $g = 10.0\text{ m/s}^2$).
(a) Determine the speed $V$ of the block and embedded bullet immediately after impact.
(b) Calculate the initial speed $u$ of the bullet.
(c) Calculate the fraction and percentage of the bullet’s initial kinetic energy converted into heat during the impact.
Solution:
(a) After the collision, the combined mass $(m + M) = 2.0\text{ kg}$ swings upward under gravity, conserving mechanical energy.
Maximum height reached:
$$h = L(1 – \cos\theta) = 2.0(1 – \cos 60^\circ) = 2.0(1 – 0.50) = 1.0\text{ meter}$$
By conservation of mechanical energy for the swing:
$$\frac{1}{2}(m + M)V^2 = (m + M)gh \implies V = \sqrt{2gh} = \sqrt{2(10.0)(1.0)} = \sqrt{20.0} \approx 4.472\text{ m/s}$$
(b) Applying conservation of linear momentum along the horizontal direction during impact:
$$m u + 0 = (m + M)V \implies u = \left(\frac{m + M}{m}\right)V = \left(\frac{0.020 + 1.98}{0.020}\right)\sqrt{20} = \left(\frac{2.0}{0.020}\right)\sqrt{20} = 100\sqrt{20} = 200\sqrt{5} \approx 447.2\text{ m/s}$$
(c) Initial kinetic energy: $K_i = \frac{1}{2}m u^2 = \frac{1}{2}(0.020)(100\sqrt{20})^2 = 0.010 \times 200,000 = 2000.0\text{ J}$.
Kinetic energy immediately after collision:
$$K_f = \frac{1}{2}(m + M)V^2 = \frac{1}{2}(2.0)(20.0) = 20.0\text{ J}$$
Mechanical energy lost to heat: $\Delta K_{\text{loss}} = 2000.0 – 20.0 = 1980.0\text{ J}$.
Fraction lost:
$$\frac{\Delta K_{\text{loss}}}{K_i} = \frac{1980}{2000} = \frac{M}{m + M} = \frac{1.98}{2.00} = 0.99 = 99.0\%$$
Insight: $99\%$ of the bullet’s kinetic energy is converted into thermal and deformation energy; only $1\%$ remains as kinetic energy of the pendulum!
Example 4 (Edge Case – Multiple Inelastic Collisions & Wall Rebound):
A block $A$ of mass $m$ slides with speed $u$ on a smooth horizontal floor toward a stationary block $B$ of mass $2m$. The coefficient of restitution between $A$ and $B$ is $e = 1/2$. A fixed rigid vertical wall is situated to the right of block $B$. Collisions with the wall are perfectly elastic ($e_w = 1$).
(a) Determine the velocities of $A$ and $B$ after their first collision.
(b) Block $B$ rebounds elastically from the wall and collides with $A$ a second time. Find the final velocities of $A$ and $B$ after this second collision.
(c) Calculate the total fractional kinetic energy lost by the system after both collisions have concluded.
Solution:
(a) First collision ($m_1 = m, u_1 = u, m_2 = 2m, u_2 = 0, e = 1/2$):
$$v_{A1} = \left(\frac{m – (1/2)(2m)}{m + 2m}\right)u = \left(\frac{m – m}{3m}\right)u = 0$$
$$v_{B1} = \left(\frac{(1 + 1/2)m}{m + 2m}\right)u = \left(\frac{1.5m}{3m}\right)u = +\frac{1}{2}u$$
Block $A$ comes momentarily to rest ($v_{A1} = 0$), and block $B$ moves toward the wall at $u/2$.
(b) Elastic rebound from wall:
Block $B$ strikes the wall and rebounds elastically with speed reversed: $u’_B = -\frac{1}{2}u$.
Second collision: Block $B$ (mass $2m$, moving left at $u/2$) collides with block $A$ (mass $m$, stationary at rest):
– Let rightward direction be positive: $u_{B2} = -u/2$, $u_{A2} = 0$.
– By linear momentum conservation:
$$2m(-u/2) + m(0) = 2m v_{B2} + m v_{A2} \implies -mu = 2m v_{B2} + m v_{A2} \implies 2v_{B2} + v_{A2} = -u$$
– Restitution equation ($e = 1/2$):
$$v_{A2} – v_{B2} = e(u_{B2} – u_{A2}) = \frac{1}{2}\left(-\frac{u}{2} – 0\right) = -\frac{u}{4} \implies v_{A2} = v_{B2} – \frac{u}{4}$$
– Substituting:
$$2v_{B2} + \left(v_{B2} – \frac{u}{4}\right) = -u \implies 3v_{B2} = -u + \frac{u}{4} = -\frac{3}{4}u \implies v_{B2} = -\frac{1}{4}u$$
$$v_{A2} = -\frac{1}{4}u – \frac{1}{4}u = -\frac{1}{2}u$$
Both blocks move to the left away from the wall: block $A$ at speed $u/2$ and block $B$ at speed $u/4$. Since $v_{A2} = -u/2$ is faster than $v_{B2} = -u/4$, they never collide again.
(c) Initial kinetic energy: $K_i = \frac{1}{2}m u^2$.
Final kinetic energy after both collisions:
$$K_f = \frac{1}{2}m v_{A2}^2 + \frac{1}{2}(2m) v_{B2}^2 = \frac{1}{2}m \left(-\frac{u}{2}\right)^2 + m \left(-\frac{u}{4}\right)^2 = \frac{1}{2}m \left(\frac{u^2}{4}\right) + m \left(\frac{u^2}{16}\right) = \frac{m u^2}{8} + \frac{m u^2}{16} = \frac{3}{16}m u^2$$
Fraction of kinetic energy retained: $\frac{K_f}{K_i} = \frac{3/16}{1/2} = \frac{3}{8}$.
Fraction of kinetic energy lost:
$$\frac{\Delta K_{\text{loss}}}{K_i} = 1 – \frac{3}{8} = \frac{5}{8} = 62.5\%$$
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
In any perfectly inelastic one-dimensional collision between two bodies ($e = 0$):
(A) Linear momentum is conserved, but mechanical kinetic energy is not conserved
(B) Kinetic energy is conserved, but linear momentum is not conserved
(C) Both linear momentum and kinetic energy are conserved
(D) Neither linear momentum nor kinetic energy is conserved
Problem 2 (JEE Main – Single Correct):
A small ball is dropped from a height $h$ onto a stationary horizontal floor. If the coefficient of restitution between the ball and floor is $e$, the maximum height to which the ball rises after $n$ successive rebounds is:
(A) $e^n h$
(B) $e^{2n} h$
(C) $e^{n/2} h$
(D) $h / e^{2n}$
Problem 3 (JEE Main – Single Correct):
A body of mass $m$ moving with speed $u$ collides head-on and sticks to an identical stationary body of mass $m$. The fraction of initial kinetic energy lost during the impact is:
(A) $1/4$
(B) $1/2$
(C) $3/4$
(D) $1$
Problem 4 (JEE Main – Single Correct):
Two bodies of masses $m_1$ and $m_2$ undergo a 1D collision with coefficient of restitution $e$. The total loss of kinetic energy is given by:
(A) $\frac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right)(u_1 – u_2)^2 (1 – e^2)$
(B) $\frac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right)(u_1 – u_2)^2 (1 – e)^2$
(C) $\frac{1}{2}(m_1 + m_2)(u_1 – u_2)^2 (1 – e^2)$
(D) $\frac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right)(u_1^2 – u_2^2)(1 – e^2)$
Problem 5 (JEE Main – Single Correct):
A body of mass $m$ moving with speed $u$ strikes an identical stationary body of mass $m$ head-on with coefficient of restitution $e$. The ratio of the velocity of the first body to that of the second body after collision is:
(A) $\frac{1 + e}{1 – e}$
(B) $\frac{1 – e}{1 + e}$
(C) $\frac{1 – e^2}{1 + e^2}$
(D) $\frac{e}{1 + e}$
Problem 6 (JEE Advanced – One or More Correct):
A particle of mass $m_1$ moving with velocity $u_1$ collides with a particle of mass $m_2$ moving with velocity $u_2$ along the same line. If the collision is partially inelastic with coefficient of restitution $e$ ($0 < e < 1$):
(A) Total linear momentum of the two-particle system is strictly conserved in any inertial frame.
(B) In the center of mass frame, the magnitude of relative velocity decreases by factor $e$.
(C) Maximum loss of kinetic energy occurs when $e = 0$.
(D) The kinetic energy associated with the center of mass motion remains completely unchanged during the collision.
Problem 7 (JEE Advanced – One or More Correct):
A ball is dropped from height $H$ onto a fixed horizontal steel floor with coefficient of restitution $e$ ($0 < e < 1$):
(A) The total distance traveled by the ball before coming to rest is $H\left(\frac{1 + e^2}{1 – e^2}\right)$.
(B) The total time elapsed before the ball comes to rest is $\sqrt{\frac{2H}{g}}\left(\frac{1 + e}{1 – e}\right)$.
(C) The theoretical number of bounces before coming to rest is infinite.
(D) The time-averaged force exerted by the ball on the floor over the entire motion equals $mg$.
Problem 8 (JEE Advanced – One or More Correct):
A bullet of mass $m$ is fired horizontally with speed $u$ into a block of mass $M$ resting on a smooth horizontal surface attached to an ideal spring of stiffness $k$. The bullet embeds completely into the block ($e = 0$):
(A) The speed of the combined mass immediately after impact is $V = \frac{m u}{m + M}$.
(B) The maximum compression of the spring is $x_{\text{max}} = u\sqrt{\frac{m^2}{(m + M)k}}$.
(C) Mechanical energy is strictly conserved during the embedding impact phase.
(D) The fraction of the bullet’s initial kinetic energy stored in the spring at maximum compression is $\frac{m}{m + M}$.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A rubber ball dropped from a height $h_0 = 16.0\text{ m}$ rebounds to a height of $9.0\text{ m}$ after its first bounce on a horizontal floor. What is the value of the coefficient of restitution $e$? (Write as a decimal, e.g., 0.75).
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A block of mass $2.0\text{ kg}$ moving at $10.0\text{ m/s}$ undergoes a head-on perfectly inelastic collision ($e = 0$) with a stationary block of mass $3.0\text{ kg}$. Calculate the total loss of kinetic energy in Joules.
Solutions & Explanations
Answer Key Summary:
1. (A) | 2. (B) | 3. (B) | 4. (A) | 5. (B) | 6. (A, B, C, D) | 7. (A, B, C, D) | 8. (A, B, D) | 9. 0.75 | 10. 60
Solution 1:
In any collision in an isolated system, external forces are zero ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$), so linear momentum is always conserved. In a perfectly inelastic collision ($e = 0$), deformation energy is not restored, so kinetic energy is not conserved.
Correct Option: (A)
Solution 2:
Speed after 1st bounce: $v_1 = e\sqrt{2gh} \implies h_1 = e^2 h$.
Speed after $n$ bounces: $v_n = e^n\sqrt{2gh} \implies h_n = (e^n)^2 h = e^{2n} h$.
Correct Option: (B)
Solution 3:
Common velocity after collision: $V = \frac{m u}{m + m} = \frac{u}{2}$.
Initial kinetic energy: $K_i = \frac{1}{2}m u^2$.
Final kinetic energy: $K_f = \frac{1}{2}(2m)V^2 = m\left(\frac{u}{2}\right)^2 = \frac{1}{4}m u^2$.
Fraction of energy lost: $\frac{K_i – K_f}{K_i} = \frac{1/2 – 1/4}{1/2} = \frac{1/4}{1/2} = \frac{1}{2}$.
Correct Option: (B)
Solution 4:
The standard formula for kinetic energy loss in a 1D collision with restitution coefficient $e$ is:
$$\Delta K_{\text{loss}} = \frac{1}{2}\left(\frac{m_1 m_2}{m_1 + m_2}\right)(u_1 – u_2)^2 (1 – e^2)$$
Correct Option: (A)
Solution 5:
For $m_1 = m_2 = m$ and $u_2 = 0$:
$$v_1 = \left(\frac{m – e m}{2m}\right)u = \left(\frac{1 – e}{2}\right)u$$
$$v_2 = \left(\frac{(1 + e)m}{2m}\right)u = \left(\frac{1 + e}{2}\right)u$$
$$\frac{v_1}{v_2} = \frac{\frac{1 – e}{2}u}{\frac{1 + e}{2}u} = \frac{1 – e}{1 + e}$$
Correct Option: (B)
Solution 6:
– (A) True: Internal forces cancel in pairs; momentum is strictly conserved in all inertial frames.
– (B) True: In the CoM frame, relative velocity magnitude is reduced from $u_{\text{rel}}$ to $e u_{\text{rel}}$.
– (C) True: $\Delta K_{\text{loss}} \propto (1 – e^2)$ is maximized when $e = 0$.
– (D) True: $K_{\text{cm}} = \frac{P_{\text{total}}^2}{2(m_1 + m_2)}$ is constant because total momentum is constant.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $D = H + 2H(e^2 + e^4 + \dots) = H\left(1 + \frac{2e^2}{1 – e^2}\right) = H\left(\frac{1 + e^2}{1 – e^2}\right)$.
– (B) True: $T = t_0 + 2t_0(e + e^2 + \dots) = t_0\left(1 + \frac{2e}{1 – e}\right) = t_0\left(\frac{1 + e}{1 – e}\right) = \sqrt{\frac{2H}{g}}\left(\frac{1 + e}{1 – e}\right)$.
– (C) True: The geometric series has infinitely many terms ($n \to \infty$) though the sum converges to a finite value.
– (D) True: Total downward momentum delivered to the floor plus upward rebound momentum over total time $T$ yields $F_{\text{avg}} = mg$.
Correct Options: (A, B, C, D)
Solution 8:
– (A) True: Horizontal momentum conservation gives $m u = (m + M)V \implies V = \frac{m u}{m + M}$.
– (B) True: Energy conservation for the spring oscillation after impact: $\frac{1}{2}k x_{\text{max}}^2 = \frac{1}{2}(m + M)V^2 \implies k x_{\text{max}}^2 = (m + M)\left(\frac{m u}{m + M}\right)^2 = \frac{m^2 u^2}{m + M} \implies x_{\text{max}} = u\sqrt{\frac{m^2}{(m + M)k}}$.
– (C) False: The embedding process is completely inelastic ($e = 0$); mechanical energy is heavily dissipated into heat.
– (D) True: Fraction of initial kinetic energy stored in spring: $\frac{\frac{1}{2}k x_{\text{max}}^2}{\frac{1}{2}m u^2} = \frac{\frac{m^2 u^2}{2(m + M)}}{\frac{1}{2}m u^2} = \frac{m}{m + M}$.
Correct Options: (A, B, D)
Solution 9:
$h_1 = e^2 h_0 \implies 9.0 = e^2 (16.0) \implies e^2 = \frac{9.0}{16.0} = \frac{9}{16} \implies e = \sqrt{\frac{9}{16}} = \frac{3}{4} = 0.75$.
Correct Answer: 0.75
Solution 10:
Given $m_1 = 2.0\text{ kg}, m_2 = 3.0\text{ kg}, u_1 = 10.0\text{ m/s}, u_2 = 0, e = 0$.
Reduced mass: $\mu = \frac{2.0 \times 3.0}{2.0 + 3.0} = \frac{6.0}{5.0} = 1.2\text{ kg}$.
Loss in kinetic energy:
$$\Delta K_{\text{loss}} = \frac{12\text{ kg}$.
Loss in kinetic energy:
$$\Delta K_{\text{loss}} = \frac{1}{2}\mu (u_1 – u_2)^2 (1 – e^2) = \frac{1}{2}(1.2)(10.0)^2 (1 – 0) = 0.6 \times 100 = 60\text{ Joules}$$.
Correct Answer: 60