At a certain height a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of `10ms^(-1)`. The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular to each other is `(g=10ms^(-2))`

Options

A. 1s
B. 2s
C. 1.5s
D. 1.75 s

Answer

To solve the problem, we need to determine the time interval after an explosion when the velocity vectors of two equal fragments become perpendicular to each other. Here’s a step-by-step solution:

### Step 1: Understand the Initial Conditions
The body is at rest before the explosion, meaning its initial velocity is zero. When it explodes, one fragment receives a horizontal velocity of \(10 \, \text{m/s}\).

### Step 2: Define the Velocities of the Fragments
Let’s denote:
- Fragment 1 (moving horizontally):
\[ \mathbf{v_1} = 10 \hat{i} \, \text{m/s} \]
- Fragment 2 (moving in the opposite horizontal direction):
\[ \mathbf{v_2} = -10 \hat{i} \, \text{m/s} \]

However, both fragments will also experience vertical motion due to gravity. The vertical velocity of each fragment after time \(t\) can be expressed as:
- For Fragment 1:
\[ \mathbf{v_{1y}} = -gt \hat{j} = -10t \hat{j} \, \text{m/s} \]
- For Fragment 2:
\[ \mathbf{v_{2y}} = -gt \hat{j} = -10t \hat{j} \, \text{m/s} \]

### Step 3: Write the Total Velocity Vectors
Now, we can express the total velocity vectors for both fragments:
- Fragment 1:
\[ \mathbf{v_1} = 10 \hat{i} - 10t \hat{j} \]
- Fragment 2:
\[ \mathbf{v_2} = -10 \hat{i} - 10t \hat{j} \]

### Step 4: Determine When the Vectors are Perpendicular
Two vectors are perpendicular when their dot product is zero:
\[ \mathbf{v_1} \cdot \mathbf{v_2} = 0 \]
Calculating the dot product:
\[ (10 \hat{i} - 10t \hat{j}) \cdot (-10 \hat{i} - 10t \hat{j}) = 0 \]
Expanding this:
\[ 10 \cdot (-10) + (-10t) \cdot (-10t) = -100 + 100t^2 = 0 \]

### Step 5: Solve for Time \(t\)
Setting the equation to zero:
\[ -100 + 100t^2 = 0 \]
\[ 100t^2 = 100 \]
\[ t^2 = 1 \]
\[ t = 1 \, \text{s} \]

### Conclusion
The time interval after the explosion for which the velocity vectors of the two fragments become perpendicular is \(1 \, \text{s}\).

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