A body of 2 kg mass makes an elastic collision with another body at rest. The velocity of the 2 kg mass is reduced to one-fourth of the original velocity. However, direction remains unchanged. The mass of the body struck is `M xx 10^(-1) kg`. Find the value of M.

Explanation

To solve the problem step by step, we will use the principles of conservation of momentum and the properties of elastic collisions.

### Step 1: Define the variables
Let:
- Mass of the first body (m1) = 2 kg
- Initial velocity of the first body (u1) = v (unknown)
- Final velocity of the first body (v1) = v/4 (given)
- Mass of the second body (m2) = M × 10^(-1) kg (unknown)
- Initial velocity of the second body (u2) = 0 (at rest)
- Final velocity of the second body (v2) = v1 (unknown)

### Step 2: Apply the conservation of momentum
The total momentum before the collision must equal the total momentum after the collision. Therefore, we can write:

\[ m_1 \cdot u_1 + m_2 \cdot u_2 = m_1 \cdot v_1 + m_2 \cdot v_2 \]

Substituting the known values:

\[ 2 \cdot v + 0 = 2 \cdot \frac{v}{4} + (M \times 10^{-1}) \cdot v_2 \]

This simplifies to:

\[ 2v = \frac{2v}{4} + (M \times 10^{-1}) \cdot v_2 \]

### Step 3: Simplify the momentum equation
Now, simplifying the equation:

\[ 2v = \frac{v}{2} + (M \times 10^{-1}) \cdot v_2 \]

Multiplying through by 2 to eliminate the fraction:

\[ 4v = v + 2(M \times 10^{-1}) \cdot v_2 \]

Subtracting v from both sides:

\[ 3v = 2(M \times 10^{-1}) \cdot v_2 \]

### Step 4: Use the elastic collision property
For elastic collisions, the relative velocity of separation is equal to the relative velocity of approach. Thus:

\[ v_2 - v_1 = -(u_1 - u_2) \]

Substituting the known values:

\[ v_2 - \frac{v}{4} = -(v - 0) \]

This simplifies to:

\[ v_2 - \frac{v}{4} = -v \]

Rearranging gives:

\[ v_2 = -v + \frac{v}{4} \]
\[ v_2 = -\frac{4v}{4} + \frac{v}{4} \]
\[ v_2 = -\frac{3v}{4} \]

### Step 5: Substitute v2 back into the momentum equation
Now we substitute \( v_2 = -\frac{3v}{4} \) back into the momentum equation:

\[ 3v = 2(M \times 10^{-1}) \cdot \left(-\frac{3v}{4}\right) \]

This simplifies to:

\[ 3v = -\frac{3}{2}(M \times 10^{-1})v \]

### Step 6: Cancel v and solve for M
Assuming \( v \neq 0 \), we can cancel \( v \):

\[ 3 = -\frac{3}{2}(M \times 10^{-1}) \]

Multiplying both sides by -2:

\[ -6 = 3M \times 10^{-1} \]

Dividing both sides by 3:

\[ -2 = M \times 10^{-1} \]

### Step 7: Solve for M
To find M, we multiply both sides by 10:

\[ M = -20 \]

However, since mass cannot be negative, we take the absolute value:

\[ M = 20 \]

### Final Answer
Thus, the value of \( M \) is 20.