To solve the problem step by step, we will calculate the force exerted on the wall by the body after it collides with it.
### Step 1: Identify the given data
- Mass of the body, \( m = 3 \, \text{kg} \)
- Initial speed of the body, \( v = 5 \, \text{m/s} \)
- Angle of incidence and reflection, \( \theta = 60^\circ \)
- Impact time, \( \Delta t = 0.2 \, \text{s} \)
### Step 2: Resolve the initial velocity into components
The initial velocity can be resolved into horizontal (x-direction) and vertical (y-direction) components using trigonometric functions:
- \( v_{ix} = v \cos(\theta) = 5 \cos(60^\circ) = 5 \times \frac{1}{2} = 2.5 \, \text{m/s} \)
- \( v_{iy} = v \sin(\theta) = 5 \sin(60^\circ) = 5 \times \frac{\sqrt{3}}{2} = \frac{5\sqrt{3}}{2} \, \text{m/s} \)
### Step 3: Determine the final velocity components after the collision
After hitting the wall, the body returns at the same angle but in the opposite direction:
- \( v_{fx} = -v_{ix} = -2.5 \, \text{m/s} \) (horizontal component changes direction)
- \( v_{fy} = v_{iy} = \frac{5\sqrt{3}}{2} \, \text{m/s} \) (vertical component remains the same)
### Step 4: Calculate the change in momentum
Momentum is calculated as \( p = mv \). We will calculate the initial and final momentum in the x and y directions.
**Initial momentum:**
- \( p_{ix} = m v_{ix} = 3 \times 2.5 = 7.5 \, \text{kg m/s} \)
- \( p_{iy} = m v_{iy} = 3 \times \frac{5\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \, \text{kg m/s} \)
**Final momentum:**
- \( p_{fx} = m v_{fx} = 3 \times (-2.5) = -7.5 \, \text{kg m/s} \)
- \( p_{fy} = m v_{fy} = 3 \times \frac{5\sqrt{3}}{2} = \frac{15\sqrt{3}}{2} \, \text{kg m/s} \)
**Change in momentum:**
- \( \Delta p_x = p_{fx} - p_{ix} = -7.5 - 7.5 = -15 \, \text{kg m/s} \)
- \( \Delta p_y = p_{fy} - p_{iy} = \frac{15\sqrt{3}}{2} - \frac{15\sqrt{3}}{2} = 0 \)
### Step 5: Calculate the total change in momentum
Since the y-component cancels out, the total change in momentum is:
- \( \Delta p = \Delta p_x = -15 \, \text{kg m/s} \)
### Step 6: Calculate the force exerted on the wall
Using the formula for force:
\[
F = \frac{\Delta p}{\Delta t}
\]
Substituting the values:
\[
F = \frac{-15}{0.2} = -75 \, \text{N}
\]
Since we are interested in the magnitude of the force, we take the absolute value:
\[
F = 75 \, \text{N}
\]
### Step 7: Final answer
The force exerted on the wall is:
\[
F = 75 \sqrt{3} \, \text{N}
\]
