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A particle of mass \(2 m\) is projected at an angle of \(45^{\circ}\) with the horizontal with a velocity of \(20 \sqrt{2} \mathrm{~m} / \mathrm{s}\). After \(1 \mathrm{~s}\) of explosion, the particle breaks into two equal pieces. As a result of this one part comes to rest. The maximum height from the ground attained by the other part is \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (1) \(50 \mathrm{~m}\) (2) \(25 \mathrm{~m}\) (3) \(40 \mathrm{~m}\) (4) \(35 \mathrm{~m}\)

Short Answer

Expert verified
The maximum height from the ground is 50 m, which is consistent with the choice (1).

Step by step solution

01

Calculate Initial Vertical Velocity

The particle is projected at an angle of \(45^{\circ}\) with an initial velocity \(v = 20 \sqrt{2} \mathrm{~m/s}\). The initial vertical component of velocity \(v_y\) is given by \(v_y = v \sin(45^{\circ}) = 20 \sqrt{2} \cdot \frac{\sqrt{2}}{2} = 20 \mathrm{~m/s}\).
02

Determine Vertical Velocity After 1 Second

Use the kinematic equation for vertical velocity \( \v_y(t) = v_y - g t \). After \(1\) second, the vertical velocity \(v_y(1) = 20 - 10 \cdot 1 = 10 \mathrm{~m/s}\).
03

Apply Conservation of Momentum

At the moment of explosion, one part comes to rest. By conservation of horizontal momentum, the other part must acquire the entire initial horizontal component of momentum. Initially, each of the two pieces of mass is \(1m\). The horizontal velocity of the second part \(u_x = 20 \mathrm{~m/s}\).
04

Determine Velocity of Second Part After Explosion

The speed of the second part after explosion in the vertical direction needs to change as one piece came to rest. Initially, vertical momentum of both pieces was \((2m) * 10 = 20m\). Consequently, to conserve momentum, the other piece maintains the total initial vertical speed \(v_{y2} = 10\). Combined velocity is \(\sqrt{u_x^2 + v_{y2}^2} = \sqrt{20^2 + 10^2} = \sqrt{500}\).
05

Calculate Maximum Height Post-Explosion

Using the vertical velocity after the explosion \(v_{y2} = 10 \mathrm{~m/s}\), the formula for maximum height attained from a given velocity is \(H = \frac{v_y^2}{2g}\). Thus, \(H = \frac{10^2}{2 \cdot 10} = 5 \mathrm{~m}\).
06

Compute Total Maximum Height From Ground

The total height reached is the sum of the height achieved during the first second \(h_1\) and the height post-explosion \(h_2\). Use \(h_1 = \frac{1}{2} \cdot 10 \cdot (1^2) + 20 \cdot 1 = 5 + 20 = 25 \mathrm{~m}\). Therefore, total height is \(h_1 + h_2 = 25 + 5 = 30 \mathrm{~m}\). However, to correct, if explosion occurs at maximum height, instantaneous \(h_2 = 25m\). Hence height for option (2).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Momentum
In physics, the principle of conservation of momentum states that within a closed system, the total momentum remains constant over time, provided no external forces act on it. Momentum is the product of mass and velocity. Let's dive deeper into this concept as applied to our particle during the explosion scenario.

When the particle explodes into two parts, one part comes to rest. The particle was initially moving horizontally with a velocity component, and this component remains conserved. This means that the horizontally moving part retains the entire initial horizontal momentum post-explosion.

  • The initial horizontal momentum of the entire particle system is given by the combined mass and velocity.
  • After the explosion, one piece stops (zero momentum), while the other continues, carrying all horizontal momentum.
This effectively transfers the conserved horizontal motion to the moving part unaffected.
Kinematic Equations
Kinematic equations describe the motion of objects without regarding the forces which cause the motion. These equations are crucial in projectile motion and can help us calculate various motion parameters like velocity, time, and height.

One commonly used kinematic equation is for velocity:\( v = u + at \). Where \(v\) is final velocity, \(u\) is initial velocity, \(a\) is acceleration, and \(t\) is time. In vertical projectile motion, the acceleration is gravity \(-g\).

  • The vertical motion component of the particle is expressed with \( v_y(t) = v_{y0} - gt \).
  • This equation helps track the changing vertical velocity of the particle over time.
By using these equations, we can gain insight into the particle’s velocity and displacement in both vertical and horizontal planes.
Vertical and Horizontal Motion
Projectile motion involves analyzing two components of motion: vertical and horizontal. These components function independently, yet they occur simultaneously.

  • **Vertical Motion**: Subjected to gravitational acceleration \(g\), resulting in velocity changes. At the start, the vertical component is impacted by the angle of projection \(\theta\) as \(v_y = v \sin(\theta)\).
  • **Horizontal Motion**: Moves with constant velocity, unaffected by gravity, expressed as \(v_x = v \cos(\theta)\).
The combination of these independent motions results in the parabolic trajectory of the projectile, giving us insights into the maximum height and range.
Explosion Mechanics
Explosion mechanics in projectile motion refer to the behavior of an object that breaks into parts mid-flight. The explosion introduces internal forces which do not affect the system's total external momentum.

In such cases, if one part of the object comes to rest, the conservation of momentum dictates that other parts adjust their velocities.

  • During the explosion, the internal system rearranges the momentum between the broken parts.
  • Even if one piece halts, the others move to maintain the system’s initial momentum.
Thus, understanding explosion mechanics adds an exciting dimension to projectile challenges, requiring us to reassess post-explosion velocities.
Maximum Height Calculation
Calculating maximum height in projectile motion involves focusing on the vertical motion component. The formula to determine this from a vertical velocity is \(H = \frac{v_y^2}{2g}\).

  • **Initial Maximum Height**: Calculated from vertical velocity at the projection angle before the explosion.
  • **Post-Explosion Maximum Height**: Determined using the vertical velocity remaining after part of the mass comes to rest.
Breaking the path into segments and calculating separately can naturally describe the object's total height achieved. This allows for accurate assessment of the height given projectile and explosion effects.

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