A particle is moving in a circular path. The acceleration and momentum of the particle at a certain moment are `a=(4 hat(i)+3hat(j))m//s^(2) and p=(8 hat(i)-6hat(j))"kg-m/s"`. The motion of the particle is

Options

A. uniform circular motion
B. accelerated circular motion
C. deaccelerated circular motion
D. we cannot say anything with a and p only

Answer

To determine the type of motion of the particle moving in a circular path, we can analyze the given acceleration and momentum vectors. Let's break down the solution step by step.

### Step 1: Identify the Given Vectors
We have:
- Acceleration vector \( \mathbf{a} = 4 \hat{i} + 3 \hat{j} \) m/s²
- Momentum vector \( \mathbf{p} = 8 \hat{i} - 6 \hat{j} \) kg·m/s

### Step 2: Calculate the Magnitude of the Acceleration
The magnitude of the acceleration \( |\mathbf{a}| \) can be calculated using the formula:
\[ |\mathbf{a}| = \sqrt{(a_x)^2 + (a_y)^2} \]
Substituting the values:
\[ |\mathbf{a}| = \sqrt{(4)^2 + (3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ m/s}^2 \]

### Step 3: Calculate the Magnitude of the Momentum
The magnitude of the momentum \( |\mathbf{p}| \) is calculated similarly:
\[ |\mathbf{p}| = \sqrt{(p_x)^2 + (p_y)^2} \]
Substituting the values:
\[ |\mathbf{p}| = \sqrt{(8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ kg·m/s} \]

### Step 4: Calculate the Dot Product of Acceleration and Momentum
The dot product \( \mathbf{a} \cdot \mathbf{p} \) is given by:
\[ \mathbf{a} \cdot \mathbf{p} = (4 \hat{i} + 3 \hat{j}) \cdot (8 \hat{i} - 6 \hat{j}) = 4 \cdot 8 + 3 \cdot (-6) = 32 - 18 = 14 \]

### Step 5: Use the Dot Product to Find the Angle
The dot product can also be expressed in terms of the magnitudes and the cosine of the angle \( \theta \) between them:
\[ \mathbf{a} \cdot \mathbf{p} = |\mathbf{a}| |\mathbf{p}| \cos \theta \]
Substituting the values we calculated:
\[ 14 = 5 \cdot 10 \cdot \cos \theta \]
\[ 14 = 50 \cos \theta \implies \cos \theta = \frac{14}{50} = \frac{7}{25} \]

### Step 6: Determine the Type of Motion
Since the acceleration has both radial and tangential components (as indicated by the non-zero dot product), the particle is undergoing accelerated circular motion. The tangential component of acceleration increases the speed of the particle, while the radial component changes its direction.

### Final Answer
The motion of the particle is **accelerated circular motion**.

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