Q.Two billiard balls A and B, each of mass 50g and moving in opposite directions with speed of $5$ m s$^{-1}$ each, collide and rebound with the same speed. If the collision lasts for $10^{-3}$ s, which of the following statements are true? (Note: more than one of the given options may be correct.)

The key idea here is the Impulse-Momentum Theorem, which states that the impulse imparted to an object is equal to the change in its momentum.

1. Calculate the change in momentum for one ball:

   Let the mass of each ball be $m = 50 \text{ g} = 0.05 \text{ kg}$.

   Let the initial velocity of ball A be $\vec{u} = +5 \text{ m s}^{-1}$.

   Since it rebounds with the same speed, its final velocity is $\vec{v} = -5 \text{ m s}^{-1}$.

   The change in momentum for ball A is $\Delta \vec{p} = m\vec{v} - m\vec{u} = m(\vec{v} - \vec{u})$.

   $\Delta \vec{p} = 0.05 \text{ kg} (-5 \text{ m s}^{-1} - (+5 \text{ m s}^{-1})) = 0.05 \text{ kg} (-10 \text{ m s}^{-1}) = -0.5 \text{ kg m s}^{-1}$.

2. Determine the impulse imparted to each ball:

   According to the Impulse-Momentum Theorem, the impulse $\vec{J}$ imparted to ball A is equal to its change in momentum:

   $\vec{J}_A = \Delta \vec{p} = -0.5 \text{ kg m s}^{-1}$.

   The magnitude of the impulse imparted to ball A is $|\vec{J}_A| = 0.5 \text{ kg m s}^{-1}$.

   Since $1 \text{ kg m s}^{-1} = 1 \text{ Ns}$, the impulse is $0.5 \text{ Ns}$.

   Due to symmetry and Newton's third law, the impulse imparted to ball B will have the same magnitude ($0.5 \text{ Ns}$) but opposite direction.

3. Calculate the average force on each ball:

   The impulse is also defined as the average force $\vec{F}_{avg}$ multiplied by the collision time $\Delta t$.

   $\vec{J} = \vec{F}_{avg} \Delta t$.

   Given $\Delta t = 10^{-3} \text{ s}$.

   For ball A, $\vec{F}_{avg, A} = \frac{\vec{J}_A}{\Delta t} = \frac{-0.5 \text{ kg m s}^{-1}}{10^{-3} \text{ s}} = -500 \text{ N}$.

   The magnitude of the average force on ball A is $500 \text{ N}$. …

When objects collide, they exert forces on each other for a short duration, causing their momenta to change. The concept that links force, time, and change in momentum is called impulse. The Impulse-Momentum Theorem states that the impulse acting on an object is equal to the change in its momentum. This theorem is crucial for analyzing collisions where forces are often large and act for very short times, making it difficult to determine the force directly. Instead, we can find the change in momentum and then relate it to the average force.

Here's how to solve the problem:

1. Identify Given Values and Convert Units:

   The mass of each billiard ball is given as $50 \text{ g}$. We must convert this to kilograms for consistency with SI units:

   $m = 50 \text{ g} = 0.05 \text{ kg}$

   The initial speed of each ball is $5 \text{ m s}^{-1}$.

   The final speed of each ball after rebounding is also $5 \text{ m s}^{-1}$.

   The duration of the collision is $\Delta t = 10^{-3} \text{ s}$.

2. Define a Coordinate System and Determine Velocities:

   Since the balls are moving in opposite directions and rebound, their velocities will change direction. Let's define the initial direction of ball A as positive.

   For ball A:

   Initial velocity, $v_{A,i} = +5 \text{ m s}^{-1}$

   Final velocity (after rebounding), $v_{A,f} = -5 \text{ m s}^{-1}$

   For ball B:

   Initial velocity, $v_{B,i} = -5 \text{ m s}^{-1}$

   Final velocity (after rebounding), $v_{B,f} = +5 \text{ m s}^{-1}$

   Watch out

   It is crucial to use velocities (vector quantities) and not just speeds (scalar quantities) when calculating momentum and impulse. A change in direction, even if speed remains constant, means a change in velocity and thus a non-zero impulse.

3. Calculate the Impulse Imparted to Each Ball:

   The impulse $J$ imparted to an object is its change in momentum, $\Delta p = m(v_f - v_i)$.

   For ball A:

   $J_A = m(v_{A,f} - v_{A,i})$

   $J_A = 0.05 \text{ kg} (-5 \text{ m s}^{-1} - (+5 \text{ m s}^{-1}))$

   $J_A = 0.05 \text{ kg} (-10 \text{ m s}^{-1})$

   $J_A = -0.5 \text{ kg m s}^{-1}$

   The magnitude of the impulse imparted to ball A is $|J_A| = 0.5 \text{ kg m s}^{-1}$.

   Note that $1 \text{ kg m s}^{-1}$ is equivalent to $1 \text{ Ns}$. So, $|J_A| = 0.5 \text{ Ns}$.

   For ball B:

   $J_B = m(v_{B,f} - v_{B,i})$

   $J_B = 0.05 \text{ kg} (+5 \text{ m s}^{-1} - (-5 \text{ m s}^{-1}))$

   $J_B = 0.05 \text{ kg} (+10 \text{ m s}^{-1})$

   $J_B = +0.5 \text{ kg m s}^{-1}$

   The magnitude of the impulse imparted to ball B is $|J_B| = 0.5 \text{ kg m s}^{-1} = 0.5 \text{ Ns}$.

   As expected by Newton's third law, the impulses on the two balls are equal in magnitude and opposite in direction.

4. Calculate the Average Force Exerted on Each Ball:

   The average force $F_{avg}$ is the impulse divided by the collision time $\Delta t$. …

Concept: Impulse-Momentum Theorem, $J=\Delta p$ and $F_{avg}=J/\Delta t$

Step 1: Convert mass and set up velocities.

$m = 50\text{ g} = 0.05$ kg. Taking ball A's initial direction as positive:

$v_{A,i}=+5$ m/s, $v_{A,f}=-5$ m/s (rebounds).

Step 2: Compute the impulse on ball A.

Magnitude $|J_A| = 0.5$ Ns. By symmetry (and Newton's third law), ball B

receives an impulse of the same magnitude, opposite sign.

Step 3: Compute the average force.

Step 4: Check each option against these numbers.

(a) claims $J=0.25$, $F=250$ N — both wrong (should be $0.5$, $500$). (b) …