Concept Card: Elastic Collisions in One Dimension
1. Physical Framework & Definitions:
A collision is an isolated event in which two or more interacting bodies exert relatively intense mutual forces over a brief time interval $\Delta t$, such that the effect of external forces during the impact can be neglected (the impulsive approximation). A collision is classified as elastic if and only if both total linear momentum and total mechanical kinetic energy of the system are strictly conserved before and after the collision.
- Conservation of Linear Momentum: Because internal contact forces are equal and opposite (Newton’s Third Law), the net external force on the system is zero ($\Sigma \vec{F}_{\text{ext}} = \vec{0}$):
$$\vec{p}_i = \vec{p}_f \iff m_1 \vec{u}_1 + m_2 \vec{u}_2 = m_1 \vec{v}_1 + m_2 \vec{v}_2$$ - Conservation of Kinetic Energy: In an ideal elastic collision, mechanical kinetic energy is temporarily converted into internal elastic strain energy during the deformation phase, and then $100\%$ converted back into kinetic energy during the restitution phase with zero dissipation into heat, sound, or permanent deformation:
$$K_i = K_f \iff \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$$ - One-Dimensional (Head-On) Impact: The centers of mass of the colliding bodies move along a single common line of impact before and after collision. All vectors can be treated as signed scalars along this axis.
2. Rigorous Derivation of 1D Elastic Collision Velocities
Consider two bodies of masses $m_1$ and $m_2$ moving with initial velocities $u_1$ and $u_2$ along the $+x$-axis, with $u_1 > u_2$ so that a collision occurs.
Step 1: Rearranging Momentum and Kinetic Energy Equations:
From momentum conservation:
$$m_1(u_1 – v_1) = m_2(v_2 – u_2) \quad \text{— (Equation 1)}$$
From kinetic energy conservation:
$$m_1(u_1^2 – v_1^2) = m_2(v_2^2 – u_2^2) \implies m_1(u_1 – v_1)(u_1 + v_1) = m_2(v_2 – u_2)(v_2 + u_2) \quad \text{— (Equation 2)}$$
Step 2: Velocity of Approach vs. Velocity of Separation:
Dividing Equation (2) by Equation (1) (assuming $u_1 \ne v_1$):
$$u_1 + v_1 = v_2 + u_2 \implies \mathbf{u_1 – u_2 = v_2 – v_1 = -(v_1 – v_2)}$$
Crucial Theorem: In any one-dimensional perfectly elastic collision, the relative velocity of approach equals the relative velocity of separation. The coefficient of restitution is strictly unity:
$$e = \frac{v_2 – v_1}{u_1 – u_2} = 1$$
Step 3: Solving for Post-Collision Velocities ($v_1$ and $v_2$):
From $v_2 = v_1 + u_1 – u_2$, substituting into Equation (1) yields the master velocity formulas:
$$\mathbf{v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)u_1 + \left(\frac{2m_2}{m_1 + m_2}\right)u_2}$$
$$\mathbf{v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 + \left(\frac{m_2 – m_1}{m_1 + m_2}\right)u_2}$$
3. High-Yield Special Cases
- Equal Masses ($m_1 = m_2 = m$):
$$v_1 = u_2, \qquad v_2 = u_1$$
The two colliding bodies completely exchange their velocities! If the target is initially stationary ($u_2 = 0$), the incident body stops dead ($v_1 = 0$) and the target takes off with the incident body’s entire initial speed ($v_2 = u_1$). - Target Initially at Rest ($u_2 = 0$):
$$v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)u_1, \qquad v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1$$- Fraction of Initial Kinetic Energy Retained by Projectile:
$$\frac{K_{1f}}{K_{1i}} = \frac{\frac{1}{2}m_1 v_1^2}{\frac{1}{2}m_1 u_1^2} = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)^2$$ - Fraction of Initial Kinetic Energy Transferred to Target:
$$\frac{\Delta K}{K_{1i}} = \frac{K_{2f}}{K_{1i}} = \frac{\frac{1}{2}m_2 v_2^2}{\frac{1}{2}m_1 u_1^2} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4 (m_1/m_2)}{\left(1 + m_1/m_2\right)^2}$$ - Condition for Maximum Energy Transfer: Maximizing $\frac{\Delta K}{K_{1i}}$ with respect to the mass ratio yields $m_1 = m_2$, producing $100\%$ kinetic energy transfer.
- Fraction of Initial Kinetic Energy Retained by Projectile:
- Massive Projectile Colliding with Light Stationary Target ($m_1 \gg m_2, u_2 = 0$):
$$\frac{m_1 – m_2}{m_1 + m_2} \approx 1, \qquad \frac{2m_1}{m_1 + m_2} \approx 2 \implies v_1 \approx u_1, \quad v_2 \approx 2u_1$$
The heavy projectile plows forward virtually unhindered, while the light target is hurled forward at twice the projectile’s speed. - Light Projectile Colliding with Massive Stationary Target ($m_1 \ll m_2, u_2 = 0$):
$$\frac{m_1 – m_2}{m_1 + m_2} \approx -1, \qquad \frac{2m_1}{m_1 + m_2} \approx 0 \implies v_1 \approx -u_1, \quad v_2 \approx 0$$
The light projectile rebounds backward with its original speed (e.g., a tennis ball striking a massive rigid wall). - Light Particle Colliding with Massive Moving Target ($m_2 \gg m_1$, Target Speed $V$):
If a heavy wall/piston moves toward a light particle at speed $V$ (so $u_2 = -V$) and the particle has initial speed $u_1 = +u$:
$$v_1 = -u_1 + 2u_2 = -u + 2(-V) = -(u + 2V)$$
The particle rebounds with speed $u + 2V$, gaining kinetic energy from the work done by the moving wall.
4. Center of Mass (CoM) Frame Formulation (JEE Advanced Method)
- Velocity of the Center of Mass: $v_{\text{cm}} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}$.
- In the CoM frame, total linear momentum is identically zero: $\vec{p}_{\text{cm}} = \vec{0}$.
- Key Theorem in CoM Frame: In a 1D elastic collision, the magnitude of each particle’s velocity relative to the center of mass remains unchanged; only the direction reverses:
$$v_{1,\text{cm}} = -u_{1,\text{cm}} \implies v_1 – v_{\text{cm}} = -(u_1 – v_{\text{cm}}) \implies \mathbf{v_1 = 2v_{\text{cm}} – u_1}$$
$$v_{2,\text{cm}} = -u_{2,\text{cm}} \implies v_2 – v_{\text{cm}} = -(u_2 – v_{\text{cm}}) \implies \mathbf{v_2 = 2v_{\text{cm}} – u_2}$$ - Maximum Elastic Strain Energy During Collision: At the instant of maximum deformation, both particles move with the common speed $v_{\text{cm}}$. The maximum potential energy stored is given by the reduced mass $\mu = \frac{m_1 m_2}{m_1 + m_2}$:
$$U_{\text{max}} = \frac{1}{2}\mu (u_1 – u_2)^2$$
5. Common JEE Pitfalls & Traps
- Trap 1 (Instantaneous vs. Asymptotic Energy Conservation): Kinetic energy is not constant during the collision. At the moment of maximum deformation, system kinetic energy reaches a minimum ($K_{\text{min}} = \frac{1}{2}(m_1 + m_2)v_{\text{cm}}^2$), with the deficit temporarily stored as internal elastic strain energy. Kinetic energy is conserved only between the initial state before collision and final state after separation.
- Trap 2 (Sign Conventions): Velocities are vector quantities. When substituting into formulas, establish a positive axis and assign negative signs to velocities directed oppositely.
- Trap 3 (Energy Transfer vs. Velocity Maximization): Maximum velocity of the target occurs when $m_1 \gg m_2$ ($v_2 \to 2u_1$), but maximum kinetic energy transfer occurs when $m_1 = m_2$ ($100\%$ transfer).
Solved Examples
Example 1 (Direct Conceptual Application – Velocity & Energy Partition):
A block of mass $m_1 = 2.0\text{ kg}$ moving with velocity $u_1 = 6.0\text{ m/s}$ along the positive $x$-axis undergoes a head-on elastic collision with a stationary block of mass $m_2 = 4.0\text{ kg}$ on a frictionless horizontal floor.
(a) Determine the velocities of both blocks after the collision.
(b) Calculate the fraction of initial kinetic energy retained by $m_1$ and transferred to $m_2$.
(c) Verify momentum and kinetic energy conservation explicitly.
Solution:
(a) Given $m_1 = 2.0\text{ kg}, u_1 = +6.0\text{ m/s}, m_2 = 4.0\text{ kg}, u_2 = 0$.
Using the velocity formulas for 1D elastic collisions:
$$v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)u_1 = \left(\frac{2.0 – 4.0}{2.0 + 4.0}\right)(6.0) = \left(\frac{-2.0}{6.0}\right)(6.0) = -2.0\text{ m/s}$$
$$v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1 = \left(\frac{2 \times 2.0}{2.0 + 4.0}\right)(6.0) = \left(\frac{4.0}{6.0}\right)(6.0) = +4.0\text{ m/s}$$
Block $m_1$ rebounds in the negative $x$-direction at $2.0\text{ m/s}$, while block $m_2$ moves forward along the positive $x$-axis at $4.0\text{ m/s}$.
(b) Initial kinetic energy of the system:
$$K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(2.0)(6.0)^2 = 36.0\text{ J}$$
Final kinetic energy of $m_1$:
$$K_{1f} = \frac{1}{2}m_1 v_1^2 = \frac{1}{2}(2.0)(-2.0)^2 = 4.0\text{ J}$$
Fraction of kinetic energy retained by $m_1$:
$$\frac{K_{1f}}{K_i} = \frac{4.0\text{ J}}{36.0\text{ J}} = \frac{1}{9} \approx 11.11\%$$
Final kinetic energy of $m_2$:
$$K_{2f} = \frac{1}{2}m_2 v_2^2 = \frac{1}{2}(4.0)(4.0)^2 = 32.0\text{ J}$$
Fraction of kinetic energy transferred to $m_2$:
$$\frac{K_{2f}}{K_i} = \frac{32.0\text{ J}}{36.0\text{ J}} = \frac{8}{9} \approx 88.89\%$$
(c) Verification:
– Total initial momentum: $p_i = m_1 u_1 + m_2 u_2 = 2.0(6.0) + 0 = +12.0\text{ kg}\cdot\text{m/s}$.
– Total final momentum: $p_f = m_1 v_1 + m_2 v_2 = 2.0(-2.0) + 4.0(4.0) = -4.0 + 16.0 = +12.0\text{ kg}\cdot\text{m/s}$ ($p_i = p_f$).
– Total final kinetic energy: $K_f = K_{1f} + K_{2f} = 4.0 + 32.0 = 36.0\text{ J} = K_i$. Both quantities are strictly conserved.
Example 2 (Mathematical Manipulation – Multiple Masses & Velocity Maximization):
Three spheres $A, B,$ and $C$ are aligned on a smooth horizontal line. Sphere $A$ (mass $m$) moves with speed $u$ and strikes stationary sphere $B$ (mass $M$) head-on and elastically. Sphere $B$ subsequently strikes stationary sphere $C$ (mass $m$) head-on and elastically.
(a) If $M = m$, determine the final velocities of all three spheres.
(b) For an arbitrary mass $M$, express the final velocity $v_C$ of sphere $C$ in terms of $u, m,$ and $M$.
(c) Find the optimal value of $M$ that maximizes the speed $v_C$, and calculate this maximum speed.
Solution:
(a) When $M = m$, sphere $A$ collides with identical stationary sphere $B$ and completely transfers its velocity: $v_A = 0, v_B = u$. Next, sphere $B$ strikes identical stationary sphere $C$: $v_B = 0, v_C = u$. Final velocities: $v_A = 0, v_B = 0, v_C = u$.
(b) For arbitrary $M$:
– First collision ($A$ of mass $m$ strikes $B$ of mass $M$ at rest):
$$v_B = \left(\frac{2m}{m + M}\right)u$$
– Second collision ($B$ of mass $M$ strikes $C$ of mass $m$ at rest):
$$v_C = \left(\frac{2M}{M + m}\right)v_B = \left(\frac{2M}{M + m}\right)\left(\frac{2m}{m + M}\right)u = \frac{4 m M}{(m + M)^2}u$$
(c) To maximize $v_C$ with respect to $M$, express the denominator:
$$\frac{4mM}{(m + M)^2} = \frac{4m}{\frac{m^2}{M} + 2m + M}$$
By the Arithmetic Mean – Geometric Mean inequality (AM $\ge$ GM):
$$\frac{m^2}{M} + M \ge 2\sqrt{\frac{m^2}{M} \times M} = 2m$$
The denominator is minimized when the two terms are equal: $\frac{m^2}{M} = M \implies M^2 = m^2 \implies M = m$.
At $M = m$:
$$v_{C,\text{max}} = \frac{4 m(m)}{(m + m)^2}u = \frac{4m^2}{4m^2}u = u$$
Key Takeaway: Maximum velocity and kinetic energy transmission through an intermediate coupling mass occurs when the masses are perfectly matched ($M = m$). This is the mechanical analogue of impedance matching.
Example 3 (Standard JEE Advanced Scenario – Elastic Collision via Spring):
A block $A$ of mass $m_1 = 1.0\text{ kg}$ moves with speed $u_1 = 12.0\text{ m/s}$ along a frictionless horizontal track toward a stationary block $B$ of mass $m_2 = 3.0\text{ kg}$. Attached to the facing front of block $B$ is a massless ideal spring of force constant $k = 300\text{ N/m}$.
(a) Determine the velocity of the center of mass of the system.
(b) Find the maximum compression $x_{\text{max}}$ of the spring during the interaction.
(c) Find the velocities of both blocks when block $A$ separates from the spring.
Solution:
(a) Since no external horizontal forces act on the system, the center of mass moves at a constant velocity:
$$v_{\text{cm}} = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} = \frac{(1.0)(12.0) + (3.0)(0)}{1.0 + 3.0} = \frac{12.0}{4.0} = 3.0\text{ m/s}$$
(b) Maximum spring compression occurs when both blocks move with the identical velocity $v_{\text{cm}} = 3.0\text{ m/s}$ (relative velocity is momentarily zero).
Initial total mechanical energy:
$$E = K_i = \frac{1}{2}m_1 u_1^2 = \frac{1}{2}(1.0)(12.0)^2 = 72.0\text{ J}$$
Kinetic energy of the system at maximum compression:
$$K_{\text{cm}} = \frac{1}{2}(m_1 + m_2)v_{\text{cm}}^2 = \frac{1}{2}(4.0)(3.0)^2 = 18.0\text{ J}$$
By conservation of mechanical energy, the deficit in kinetic energy is stored entirely as spring potential energy:
$$\frac{1}{2}k x_{\text{max}}^2 = K_i – K_{\text{cm}} = 72.0 – 18.0 = 54.0\text{ J}$$
$$\frac{1}{2}(300) x_{\text{max}}^2 = 54.0 \implies 150 x_{\text{max}}^2 = 54.0 \implies x_{\text{max}}^2 = \frac{54.0}{150} = 0.36\text{ m}^2$$
$$x_{\text{max}} = \sqrt{0.36} = 0.60\text{ m} = 60.0\text{ cm}$$
Alternative calculation via reduced mass:
$$\mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{1.0 \times 3.0}{4.0} = 0.75\text{ kg}$$
$$\frac{1}{2}k x_{\text{max}}^2 = \frac{1}{2}\mu (u_1 – u_2)^2 = \frac{1}{2}(0.75)(12.0)^2 = 54.0\text{ J} \implies x_{\text{max}} = 0.60\text{ m}$$.
(c) Since the spring is perfectly elastic and releases all stored energy without loss, the entire process is a 1D elastic collision. Using the Center of Mass reversal method:
$$v_1 = 2v_{\text{cm}} – u_1 = 2(3.0) – 12.0 = 6.0 – 12.0 = -6.0\text{ m/s}$$
$$v_2 = 2v_{\text{cm}} – u_2 = 2(3.0) – 0 = +6.0\text{ m/s}$$
Block $A$ rebounds to the left at $6.0\text{ m/s}$, and block $B$ moves to the right at $6.0\text{ m/s}$.
Example 4 (Edge Case – Collision with Moving Massive Wall):
A small elastic ball of mass $m = 0.20\text{ kg}$ is traveling horizontally at $u = 10.0\text{ m/s}$ toward a massive vertical wall of mass $M \gg m$. The wall is moving toward the ball at a steady speed of $V = 3.0\text{ m/s}$. The collision is head-on and perfectly elastic, lasting for $\Delta t = 2.0\text{ ms}$.
(a) Find the speed of the ball immediately after rebounding.
(b) Determine the average force exerted by the wall on the ball during the impact.
(c) Calculate the gain in kinetic energy of the ball, and explain physically where this energy originates.
Solution:
(a) Let the initial direction of motion of the ball be $+x$.
Initial velocity of ball: $u_1 = +10.0\text{ m/s}$.
Initial velocity of wall: $u_2 = -3.0\text{ m/s}$.
Relative velocity of approach: $v_{\text{rel, app}} = u_1 – u_2 = 10.0 – (-3.0) = 13.0\text{ m/s}$.
For an elastic collision ($e = 1$), the relative velocity of separation must equal the velocity of approach:
$$v_2 – v_1 = 13.0\text{ m/s}$$
Because the wall is extraordinarily massive ($M \gg m$), its velocity is unaffected by the tiny ball: $v_2 = u_2 = -3.0\text{ m/s}$.
$$-3.0 – v_1 = 13.0 \implies v_1 = -16.0\text{ m/s}$$
The ball rebounds with a speed of $16.0\text{ m/s}$ (in general, $v_{\text{rebound}} = u + 2V$).
(b) Change in linear momentum of the ball:
$$\Delta p = m(v_1 – u_1) = 0.20(-16.0 – 10.0) = 0.20(-26.0) = -5.2\text{ kg}\cdot\text{m/s}$$
Average force exerted on the ball:
$$F_{\text{avg}} = \frac{|\Delta p|}{\Delta t} = \frac{5.2\text{ N}\cdot\text{s}}{2.0 \times 10^{-3}\text{ s}} = 2600\text{ N}$$
(c) Initial kinetic energy of the ball: $K_i = \frac{1}{2}(0.20)(10.0)^2 = 10.0\text{ J}$.
Final kinetic energy of the ball: $K_f = \frac{1}{2}(0.20)(16.0)^2 = 25.6\text{ J}$.
$$\Delta K = K_f – K_i = 25.6 – 10.0 = +15.6\text{ J}$$
Physical Origin: During the contact duration $\Delta t$, the wall continues to move forward into the ball. The normal contact force exerted by the wall acts in the same direction as the wall’s displacement ($W = \int \vec{F}_{\text{normal}} \cdot d\vec{r}_{\text{wall}} > 0$). Positive mechanical work is performed on the ball by the moving wall, supplied by the external mechanism or motor maintaining the wall’s uniform velocity.
Worksheet: 10 Practice Problems
Problem 1 (JEE Main – Single Correct):
A particle of mass $m_1$ moving with initial velocity $u$ undergoes a head-on elastic collision with a stationary particle of mass $m_2$. If the incident particle $m_1$ comes completely to rest after the collision, the ratio of masses $m_1/m_2$ must be:
(A) $1/2$
(B) $1$
(C) $2$
(D) $4$
Problem 2 (JEE Main – Single Correct):
A body of mass $m$ moving with speed $v$ collides head-on and elastically with a stationary body of mass $2m$. The fraction of initial kinetic energy transferred to the stationary body is:
(A) $1/9$
(B) $4/9$
(C) $8/9$
(D) $1/3$
Problem 3 (JEE Main – Single Correct):
In any one-dimensional perfectly elastic collision between two isolated bodies, the relative velocity of separation after the collision is:
(A) Greater than the relative velocity of approach
(B) Equal in magnitude to the relative velocity of approach
(C) Less than the relative velocity of approach
(D) Identically zero
Problem 4 (JEE Main – Single Correct):
A small steel sphere is dropped from a height $h_0$ onto a fixed, massive, perfectly elastic horizontal floor. If the coefficient of restitution is $e = 1$, the height to which the sphere rebounds after the first impact is:
(A) $h_0/2$
(B) $h_0$
(C) $2h_0$
(D) $h_0/4$
Problem 5 (JEE Main – Single Correct):
In a nuclear reactor moderator, a fast neutron of mass $m$ collides head-on and elastically with a stationary carbon nucleus of mass $12m$. The percentage of the neutron’s initial kinetic energy transferred to the carbon nucleus is approximately:
(A) $28.4\%$
(B) $71.6\%$
(C) $50.0\%$
(D) $92.3\%$
Problem 6 (JEE Advanced – One or More Correct):
Two bodies $A$ and $B$ undergo a one-dimensional head-on elastic collision. Which of the following statements is/are correct?
(A) The total linear momentum of the two-body system is conserved throughout the entire collision.
(B) Total kinetic energy is conserved before and after the collision, but is not constant during the deformation interval.
(C) In the center of mass reference frame, the velocities of both particles are reversed in direction without any change in magnitude.
(D) If the colliding bodies have equal masses, they strictly exchange their velocities upon collision.
Problem 7 (JEE Advanced – One or More Correct):
A projectile of mass $m_1$ moving with velocity $u$ strikes a stationary target of mass $m_2$ head-on in an elastic collision. The fraction of initial kinetic energy transferred to the target is $\eta$. Which of the following statements is/are correct?
(A) $\eta = \frac{4 m_1 m_2}{(m_1 + m_2)^2}$.
(B) $\eta$ is maximum when $m_1 = m_2$, with a maximum value of $\eta_{\text{max}} = 1$.
(C) The target receives the same fraction of kinetic energy whether the mass ratio $m_1/m_2$ is equal to $r$ or to $1/r$.
(D) The final speed of the target is maximum when $m_1 = m_2$.
Problem 8 (JEE Advanced – One or More Correct):
A block of mass $m$ moving with speed $u$ along $+x$ on a frictionless floor collides head-on elastically with a stationary block of mass $M$ equipped with a light spring of constant $k$. At the instant of maximum compression of the spring:
(A) Both blocks move with a common velocity equal to $\frac{m u}{m + M}$.
(B) The kinetic energy of the system is at its minimum value, equal to $\frac{1}{2}\frac{m^2 u^2}{m + M}$.
(C) The maximum compression of the spring is $x_{\text{max}} = u\sqrt{\frac{m M}{(m + M)k}}$.
(D) The acceleration of the center of mass of the two-block system is zero throughout the interaction.
Problem 9 (JEE Main / Advanced – Numerical Value Type):
A block of mass $m_1 = 4.0\text{ kg}$ moving at $10.0\text{ m/s}$ along the $+x$-axis collides elastically and head-on with a stationary block of mass $m_2 = 6.0\text{ kg}$. Calculate the speed of the second block after the collision in $\text{m/s}$.
Problem 10 (JEE Main / Advanced – Numerical Value Type):
A small elastic pellet of mass $m$ moving horizontally at $15.0\text{ m/s}$ collides head-on and elastically with a massive heavy plate ($M \gg m$) moving toward the pellet at a steady speed of $5.0\text{ m/s}$. The speed with which the pellet rebounds from the plate in $\text{m/s}$ is:
Solutions & Explanations
Answer Key Summary:
1. (B) | 2. (C) | 3. (B) | 4. (B) | 5. (A) | 6. (A, B, C, D) | 7. (A, B, C) | 8. (A, B, C, D) | 9. 8 | 10. 25
Solution 1:
For a head-on elastic collision with a stationary target ($u_2 = 0$):
$$v_1 = \left(\frac{m_1 – m_2}{m_1 + m_2}\right)u$$
For the incident body to come completely to rest ($v_1 = 0$):
$$m_1 – m_2 = 0 \implies m_1 = m_2 \implies \frac{m_1}{m_2} = 1$$.
Correct Option: (B)
Solution 2:
The fraction of initial kinetic energy transferred to the target mass $m_2 = 2m$ (initially at rest) by incident mass $m_1 = m$ is:
$$\frac{\Delta K}{K_i} = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4(m)(2m)}{(m + 2m)^2} = \frac{8m^2}{9m^2} = \frac{8}{9}$$.
Correct Option: (C)
Solution 3:
By definition of a 1D elastic collision, kinetic energy and momentum conservation combine to yield:
$$u_1 – u_2 = v_2 – v_1 \implies |v_{\text{sep}}| = |v_{\text{app}}| \implies e = 1$$.
The relative velocity of separation is equal in magnitude to the relative velocity of approach.
Correct Option: (B)
Solution 4:
Velocity just before hitting the floor from height $h_0$ is $u = \sqrt{2g h_0}$.
Rebound speed is $v = e u = (1)\sqrt{2g h_0} = \sqrt{2g h_0}$.
Rebound height is $h_1 = \frac{v^2}{2g} = \frac{2g h_0}{2g} = h_0$.
Correct Option: (B)
Solution 5:
Here $m_1 = m$ (neutron) and $m_2 = 12m$ (carbon nucleus, $u_2 = 0$).
Fraction of kinetic energy transferred to carbon:
$$\eta = \frac{4 m_1 m_2}{(m_1 + m_2)^2} = \frac{4(1)(12)}{(1 + 12)^2} = \frac{48}{169} \approx 0.28402 = 28.4\%$$.
(The remaining $71.6\%$ is retained by the neutron).
Correct Option: (A)
Solution 6:
– (A) True: Since no external forces act on the system, momentum is conserved at every instant of time.
– (B) True: During maximum compression/deformation, kinetic energy is minimized; it is restored only when the bodies fully separate.
– (C) True: In the zero-momentum CoM frame, conservation of both momentum and kinetic energy requires $\vec{v}_{1,\text{cm}} = -\vec{u}_{1,\text{cm}}$ and $\vec{v}_{2,\text{cm}} = -\vec{u}_{2,\text{cm}}$.
– (D) True: For $m_1 = m_2$, the formulas yield $v_1 = u_2$ and $v_2 = u_1$.
Correct Options: (A, B, C, D)
Solution 7:
– (A) True: $\eta = \frac{K_{2f}}{K_{1i}} = \frac{4 m_1 m_2}{(m_1 + m_2)^2}$.
– (B) True: $\frac{d\eta}{d(m_1/m_2)} = 0 \implies m_1 = m_2$, giving $\eta_{\text{max}} = 1$ ($100\%$ energy transfer).
– (C) True: Let $r = m_1/m_2$. Then $\eta = \frac{4r}{(1 + r)^2}$. If we replace $r$ with $1/r$: $\frac{4(1/r)}{(1 + 1/r)^2} = \frac{4/r}{(r + 1)^2/r^2} = \frac{4r}{(1 + r)^2} = \eta$. Thus, $\eta$ is invariant under reciprocity of mass ratio.
– (D) False: The speed of the target is $v_2 = \frac{2m_1}{m_1 + m_2}u = \frac{2}{1 + (m_2/m_1)}u$. As $m_1/m_2 \to \infty$ ($m_1 \gg m_2$), $v_2 \to 2u$, which exceeds the value $u$ obtained when $m_1 = m_2$.
Correct Options: (A, B, C)
Solution 8:
– (A) True: At maximum compression, relative velocity is zero, so both blocks move at the system center-of-mass velocity $v_{\text{cm}} = \frac{m u}{m + M}$.
– (B) True: Kinetic energy at this moment is $K_{\text{cm}} = \frac{1}{2}(m + M)v_{\text{cm}}^2 = \frac{1}{2}(m + M)\left(\frac{m u}{m + M}\right)^2 = \frac{1}{2}\frac{m^2 u^2}{m + M}$.
– (C) True: By energy conservation: $\frac{1}{2}k x_{\text{max}}^2 = K_i – K_{\text{cm}} = \frac{1}{2}m u^2 – \frac{1}{2}\frac{m^2 u^2}{m + M} = \frac{1}{2}\frac{m M}{m + M}u^2 \implies x_{\text{max}} = u\sqrt{\frac{m M}{(m + M)k}}$.
– (D) True: Net external force is zero, so $a_{\text{cm}} = 0$ throughout.
Correct Options: (A, B, C, D)
Solution 9:
Formula for the final speed of a stationary target ($u_2 = 0$) struck elastically by $m_1$:
$$v_2 = \left(\frac{2m_1}{m_1 + m_2}\right)u_1$$
Given $m_1 = 4.0\text{ kg}, m_2 = 6.0\text{ kg}, u_1 = 10.0\text{ m/s}$:
$$v_2 = \left(\frac{2 \times 4.0}{4.0 + 6.0}\right)(10.0) = \left(\frac{8.0}{10.0}\right)(10.0) = 8.0\text{ m/s}$$.
Correct Answer: 8
Solution 10:
Let the initial direction of the pellet be positive ($+x$).
$u_1 = +15.0\text{ m/s}$ and $u_2 = -5.0\text{ m/s}$ (wall moving toward pellet).
Relative velocity of approach: $v_{\text{app}} = u_1 – u_2 = 15.0 – (-5.0) = 20.0\text{ m/s}$.
For an elastic collision with an infinitely massive plate ($M \gg m$), the plate’s speed is unaffected: $v_2 = -5.0\text{ m/s}$.
Relative velocity of separation must equal approach velocity:
$$v_2 – v_1 = 20.0\text{ m/s} \implies -5.0 – v_1 = 20.0 \implies v_1 = -25.0\text{ m/s}$$
The pellet rebounds with a speed of $25.0\text{ m/s}$. (Direct formula: $v_{\text{rebound}} = u + 2V = 15.0 + 2(5.0) = 25.0\text{ m/s}$).
Correct Answer: 25