To solve this problem, we need to analyze the motion of the particle before and after the explosion. Let's break it down step by step.
Initial Conditions
The particle has a mass of 2m and is projected at an angle of 45 degrees with an initial velocity of 2√2 m/s. We can determine the initial velocity components:
- Horizontal Component (Vx): Vx = V * cos(θ) = 2√2 * cos(45°) = 2√2 * (1/√2) = 2 m/s
- Vertical Component (Vy): Vy = V * sin(θ) = 2√2 * sin(45°) = 2√2 * (1/√2) = 2 m/s
Position After 1 Second
Next, we calculate the position of the particle after 1 second. The horizontal and vertical displacements can be calculated using the equations of motion:
- Horizontal Displacement (x): x = Vx * t = 2 m/s * 1 s = 2 m
- Vertical Displacement (y): y = Vy * t - (1/2) * g * t² = 2 m/s * 1 s - (1/2) * 9.8 m/s² * (1 s)² = 2 m - 4.9 m = -2.9 m
After 1 second, the particle is at the coordinates (2 m, -2.9 m) in the vertical direction. However, since the vertical position cannot be negative in this context, we will consider the height above the launch point, which is 2 m - 4.9 m = -2.9 m, indicating it has fallen below the launch level.
Explosion and Momentum Conservation
At the moment of the explosion, the particle breaks into two equal pieces, each with a mass of m. One piece comes to rest, meaning its velocity becomes 0. By the conservation of momentum, the momentum before the explosion must equal the momentum after the explosion.
The initial momentum of the entire particle is:
Initial Momentum = (2m) * (2 m/s) = 4m kg·m/s (in the horizontal direction)
After the explosion, one piece is at rest, so all the momentum must be carried by the other piece:
Let the velocity of the moving piece be V. Then:
m * V = 4m
V = 4 m/s
Vertical Motion of the Moving Piece
Now, we need to find the maximum height attained by the moving piece after the explosion. The vertical component of the velocity just before the explosion is still 2 m/s. After the explosion, the moving piece has a horizontal velocity of 4 m/s and retains the vertical velocity of 2 m/s.
To find the maximum height, we can use the following kinematic equation:
Vf² = Vi² + 2a * d
At the maximum height, the final vertical velocity (Vf) will be 0. The initial vertical velocity (Vi) is 2 m/s, and the acceleration (a) is -g (which is -9.8 m/s²).
Setting up the equation:
0 = (2 m/s)² + 2 * (-9.8 m/s²) * d
0 = 4 - 19.6d
19.6d = 4
d = 4 / 19.6 ≈ 0.204 m
Maximum Height Calculation
The maximum height attained by the moving piece is the height it reached after the explosion plus the height it was at just before the explosion:
Maximum Height = Initial Height + d = -2.9 m + 0.204 m ≈ -2.696 m
Since the height is measured from the launch point, the maximum height above the launch point is approximately 0.204 m. Thus, the maximum height attained by the other part after the explosion is about 0.204 m above the original launch height.
