An alpha-particle of mass m suffers `1`-dimentinal eleastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing `64%` of its initial kinetic energy. The mass of the nucleus is :

Options

A. 2 m
B. 3.5 m
C. 1.5 m
D. 4 m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have an alpha particle of mass \( m \) colliding elastically with a nucleus of unknown mass \( M \) that is initially at rest. After the collision, the alpha particle is scattered directly backward and loses 64% of its initial kinetic energy.

### Step 2: Determine the initial kinetic energy of the alpha particle
The initial kinetic energy \( KE_i \) of the alpha particle is given by:
\[ KE_i = \frac{1}{2} m u^2 \]
where \( u \) is the initial velocity of the alpha particle.

### Step 3: Calculate the final kinetic energy of the alpha particle
Since the alpha particle loses 64% of its kinetic energy, it retains 36% of its initial kinetic energy:
\[ KE_f = 0.36 \times KE_i = 0.36 \times \frac{1}{2} m u^2 \]

### Step 4: Express the final kinetic energy in terms of the final velocity
Let \( v_1 \) be the final velocity of the alpha particle after the collision. The final kinetic energy can also be expressed as:
\[ KE_f = \frac{1}{2} m v_1^2 \]
Setting the two expressions for \( KE_f \) equal gives:
\[ \frac{1}{2} m v_1^2 = 0.36 \times \frac{1}{2} m u^2 \]
We can cancel \( \frac{1}{2} m \) from both sides (assuming \( m \neq 0 \)):
\[ v_1^2 = 0.36 u^2 \]
Taking the square root:
\[ v_1 = 0.6 u \]

### Step 5: Apply conservation of momentum
Before the collision, the total momentum is:
\[ p_i = mu + 0 = mu \]
After the collision, the momentum is:
\[ p_f = -m v_1 + M v_2 \]
where \( v_2 \) is the velocity of the nucleus after the collision. Setting the initial momentum equal to the final momentum gives:
\[ mu = -m(0.6u) + Mv_2 \]
This simplifies to:
\[ mu = -0.6mu + Mv_2 \]
Rearranging gives:
\[ mu + 0.6mu = Mv_2 \]
\[ 1.6mu = Mv_2 \]

### Step 6: Use the coefficient of restitution
For elastic collisions, the coefficient of restitution \( e = 1 \) gives us:
\[ v_2 + v_1 = u \]
Substituting \( v_1 = 0.6u \):
\[ v_2 + 0.6u = u \]
Solving for \( v_2 \):
\[ v_2 = u - 0.6u = 0.4u \]

### Step 7: Substitute \( v_2 \) back into the momentum equation
Now we substitute \( v_2 = 0.4u \) back into the momentum equation:
\[ 1.6mu = M(0.4u) \]
Cancelling \( u \) (assuming \( u \neq 0 \)):
\[ 1.6m = 0.4M \]
Solving for \( M \):
\[ M = \frac{1.6m}{0.4} = 4m \]

### Conclusion
The mass of the nucleus is \( 4m \).