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A particle strikes a horizontal smooth floor with a velocity \(u\) making an angle \(\theta\) with the floor and rebounds with velocity \(v\) making an angle \(\phi\) with the floor. If the coefficient of restitution between the particle and the floor is \(e\), then (1) the impulse delivered by the floor to the body is $$m u(1+e) \sin \theta$$ (2) \(\tan \phi=e \tan \theta\) (3) \(v=u \sqrt{1-(1-e)^{2} \sin ^{2} \theta}\) (4) the ratio of final kinetic energy to the initial kinetic energy is \(\left(\cos ^{2} \theta+e^{2} \sin ^{2} \theta\right)\)

Short Answer

Expert verified
All statements (1) to (4) are correct based on the given conditions and calculations.

Step by step solution

01

Analyze Initial and Rebounded Velocities

A particle strikes the floor with initial velocity \( u \), at an angle \( \theta \) to the horizontal. This implies the initial velocity can be broken into horizontal and vertical components: \( u_x = u \cos \theta \) and \( u_y = u \sin \theta \). After rebounding, its velocity is \( v \), making an angle \( \phi \) with the floor. Similarly, the rebounded velocities are \( v_x = v \cos \phi \) and \( v_y = v \sin \phi \).
02

Use Coefficient of Restitution Formula

The coefficient of restitution \( e \) relates the velocities before and after impact along the normal (vertical) direction. According to the restitution definition, \( e = \frac{v_y}{u_y} \), so \( v \sin \phi = e u \sin \theta \). Rearranging gives \( \tan \phi = e \tan \theta \), which matches statement (2).
03

Calculate Impulse Delivered by the Floor

Impulse is the change in vertical momentum. The initial vertical component of momentum is \( mu \sin \theta \) and the final is \( -mv \sin \phi \). Thus, the impulse is \( m\left( v \sin \phi + u \sin \theta \right) = m( 1 + e) u \sin \theta \), which matches statement (1).
04

Calculate Final Velocity Magnitude

From the horizontal component, since there's no change along the floor (conservation of horizontal momentum), \( u \cos \theta = v \cos \phi \). Using previously found \( \tan \phi \), \( v \cos \phi = u \cos \theta \). Combine with vertical to isolate \( v = u \sqrt{1 - (1-e)^2 \sin^2 \theta} \), matching statement (3).
05

Compute Kinetic Energy Ratio

Initial kinetic energy = \( \frac{1}{2} m u^2 \) and final kinetic energy = \( \frac{1}{2} m v^2 \). Ratio: \( \frac{v^2}{u^2} = \left( \cos^2 \theta + e^2 \sin^2 \theta \right) \), matching statement (4).

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Key Concepts

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

Impulse and Momentum
Impulse and momentum are key concepts when analyzing the collision and rebound of a particle. When a particle impacts a surface, how it behaves after the collision depends on the change in momentum. Momentum is the product of mass and velocity, and it is a vector quantity. Here, we'll break down how impulse, a force applied over time, causes a change in momentum.

The impulse delivered by the floor to a particle can be calculated using the change in the vertical component of momentum. Initially, the vertical momentum of the particle is given by the product of its mass (\( m \)) and its vertical component velocity (\( u \sin \theta \)). After rebounding, the final vertical momentum becomes (\(-mv \sin \phi \)). The negative sign indicates that the direction has reversed.

Thus, the net change in momentum, which is the impulse, becomes:
  • \[ \text{Impulse} = m(v \sin \phi + u \sin \theta) \]
This formula can be further simplified to obtain the impulse in terms of the coefficient of restitution (\( e \)):
  • \[ \text{Impulse} = m u(1+e) \sin \theta \].
This impulse reflects the interaction between the particle and the surface during the collision and plays a crucial role in understanding how the particle moves after impact.
Kinetic Energy
When it comes to moving particles, kinetic energy is an essential concept to grasp. Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula:
  • \[ \text{Kinetic Energy} = \frac{1}{2} m v^2 \]
Initially, the particle has kinetic energy based on its velocity \( u \). After rebounding from the floor, the velocity changes to \( v \), and so does the kinetic energy. Understanding the kinetic energy ratio before and after collision helps in comprehending energy transformations during the event.

In this case, the ratio of the final kinetic energy to the initial kinetic energy can be derived by comparing the squares of the velocities, given as:
  • \[ \frac{v^2}{u^2} = \left( \cos^2 \theta + e^2 \sin^2 \theta \right) \].
This expression implies that not all kinetic energy is lost during the collision. Part of it is retained, which is influenced by the coefficient of restitution \( e \). The greater the value of \( e \), indicating a more elastic collision, the more kinetic energy remains with the particle post-collision.
Angle of Rebound
The angle at which a particle rebounds off a surface is not just a random occurrence but a result of several physical factors. The angle of rebound reflects how the velocity of a particle changes after it has impacted another surface, influenced significantly by the coefficient of restitution (\( e \)).

When the particle collides with the floor, its initial direction is characterized by the angle \( \theta \) and velocity components \( u \cos \theta \) and \( u \sin \theta \). Upon impact, the normal component of the velocity undergoes a change defined by:
  • \[ \tan \phi = e \tan \theta \]
This implies that the rebound angle \( \phi \) is altered in proportion to the restitution coefficient \( e \).

In essence, \( e \) quantifies the elasticity of the collision. A higher \( e \) leads to a steeper rebound angle (closer to the incident angle), indicating minimal energy loss. This relationship between the angles showcases the predictability of rebound behavior under specific conditions and allows us to calculate the shift in direction that occurs upon strikes with surfaces.

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