Two identical balls each of mass 4 kg are moving towards each other with speeds 2 m/s and 3 m/s respectively. They undergo head on perfectly elastic collision. Then impulse imparted by one ball on other is

Options

A. 12 Ns
B. 8 Ns
C. 20 Ns
D. 40 Ns

Answer

To solve the problem of finding the impulse imparted by one ball on the other during a perfectly elastic collision, we can follow these steps:

### Step 1: Understand the Problem
We have two identical balls, each with a mass of 4 kg. One ball is moving towards the right with a speed of 2 m/s, and the other is moving towards the left with a speed of 3 m/s. We need to find the impulse imparted by one ball on the other after they collide elastically.

### Step 2: Set Up the Conservation of Momentum
In a perfectly elastic collision, the total momentum before the collision is equal to the total momentum after the collision.

Let:
- Mass of ball 1 (m1) = 4 kg, initial velocity (u1) = 2 m/s
- Mass of ball 2 (m2) = 4 kg, initial velocity (u2) = -3 m/s (negative because it's moving in the opposite direction)

The total initial momentum (p_initial) can be calculated as:
\[ p_{\text{initial}} = m_1 \cdot u_1 + m_2 \cdot u_2 \]
\[ p_{\text{initial}} = 4 \cdot 2 + 4 \cdot (-3) \]
\[ p_{\text{initial}} = 8 - 12 = -4 \, \text{kg m/s} \]

### Step 3: Set Up the Conservation of Kinetic Energy
In a perfectly elastic collision, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

The total initial kinetic energy (KE_initial) can be calculated as:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]
\[ KE_{\text{initial}} = \frac{1}{2} \cdot 4 \cdot (2^2) + \frac{1}{2} \cdot 4 \cdot (-3)^2 \]
\[ KE_{\text{initial}} = 4 \cdot 2 + 4 \cdot 4.5 = 8 + 18 = 26 \, \text{J} \]

### Step 4: Use the Equations for Elastic Collisions
For elastic collisions, we can use the following equations:
1. \( v_1 + v_2 = u_1 + u_2 \) (conservation of momentum)
2. \( v_2 - v_1 = -(u_2 - u_1) \) (relative velocity)

Let \( v_1 \) and \( v_2 \) be the final velocities of ball 1 and ball 2, respectively.

From the first equation:
\[ v_1 + v_2 = 2 - 3 = -1 \quad (1) \]

From the second equation:
\[ v_2 - v_1 = -(-3 - 2) = 5 \quad (2) \]

### Step 5: Solve the Equations
Now we have two equations:
1. \( v_1 + v_2 = -1 \)
2. \( v_2 - v_1 = 5 \)

From equation (1):
\[ v_2 = -1 - v_1 \]

Substituting into equation (2):
\[ (-1 - v_1) - v_1 = 5 \]
\[ -1 - 2v_1 = 5 \]
\[ -2v_1 = 6 \]
\[ v_1 = -3 \, \text{m/s} \]

Now substituting back to find \( v_2 \):
\[ v_2 = -1 - (-3) = 2 \, \text{m/s} \]

### Step 6: Calculate the Impulse
Impulse (J) is defined as the change in momentum. For ball 2:
\[ J = m_2(v_2 - u_2) \]
\[ J = 4(2 - (-3)) \]
\[ J = 4(2 + 3) = 4 \cdot 5 = 20 \, \text{Ns} \]

### Final Answer
The impulse imparted by one ball on the other is **20 Ns**.

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