This question has statement-I and statement‑II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as then f \textrm{ } \left(\right. \frac{1}{2} \left(mv\right)^{2} \left.\right). Statement-II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Engineering

Physics

Newtons Second Law of Motion

Collision

Momentum and Energy Conservation for System of Particles

Question

This question has statement-I and statement‑II. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement-I: A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as then $f (\frac{1}{2} {mv}^{2})$ .

Statement-II: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

Statement-I is true, Statement-II is true, Statement-II is a correct explanation of Statement-I.

Statement-I is true, Statement-II is false

Statement-I is false, Statement-II is true

Statement-I is true, Statement-II is true, Statement-II is not a correct explanation of Statement-I

$ΔE = \frac{1}{2} µ (v_{rel}^{2} - u_{rel}^{2}) = \frac{1}{2} \frac{mM}{m + M} (u_{rel}^{2})$

This question involves analyzing two statements about inelastic collisions and determining their truth and relationship.

Concept Explanation: Inelastic Collisions and Maximum Energy Loss

In a collision between two particles, the total momentum is always conserved (if no external force acts), but kinetic energy may not be conserved. When kinetic energy is not conserved, the collision is inelastic.

Statement-II claims: "Maximum energy loss occurs when the particles get stuck together as a result of the collision." This is true. When two particles stick together after collision, it is called a perfectly inelastic collision. In this case, the loss of kinetic energy is maximum for given masses and initial conditions.

Now, let's verify Statement-I. It states that for a point particle of mass $m$ moving with speed $v$ colliding with a stationary particle of mass $M$ , the maximum energy loss possible is $f (\frac{1}{2} {mv}^{2})$ . We need to derive this maximum loss.

Step 1: Conservation of Momentum

Initial momentum = $m v$ (since mass M is stationary).

Let $v_{f}$ be the common velocity after perfectly inelastic collision (they stick together). Final mass = $m + M$ .

By conservation of momentum: $m v = (m + M) v_{f}$

So, $v_{f} = \frac{m v}{m + M}$

Step 2: Calculate Kinetic Energy Loss

Initial kinetic energy, $K_{i} = \frac{1}{2} m v^{2}$ (since M is stationary).

Final kinetic energy, $K_{f} = \frac{1}{2} (m + M) {v_{f}}^{2} = \frac{1}{2} (m + M) {(\frac{m v}{m + M})}^{2} = \frac{1}{2} \frac{m^{2}}{v}$

Energy loss, $Δ K = K_{i} - K_{f} = \frac{1}{2} m v^{2} - \frac{1}{2} \frac{m^{2}}{v} = \frac{1}{2} m v^{2} [1 - \frac{m}{m + M}] = \frac{1}{2} m v^{2} (\frac{M}{m + M})$

So, maximum energy loss is $Δ K = (\frac{M}{m + M}) (\frac{1}{2} m v^{2})$ . This is indeed of the form $f (\frac{1}{2} {mv}^{2})$ , where $f = \frac{M}{m + M}$ . So Statement-I is true.

Step 3: Relationship between Statements

Statement-II correctly explains why the energy loss is maximum (because they stick together), and we used this condition to derive the expression in Statement-I. Therefore, Statement-II is a correct explanation of Statement-I.

Final Answer: Statement-I is true, Statement-II is true, Statement-II is a correct explanation of Statement-I.

Detailed Solution

Concept Explanation: Inelastic Collisions and Maximum Energy Loss

Related Topics and Formulae

Detailed Solution

Concept Explanation: Inelastic Collisions and Maximum Energy Loss

Related Topics and Formulae

Student who searched for similar questions

Questions 1

Questions 2

Questions 3

Questions 4

Questions 5