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Chapter 9: Problem 60

(III) For an elastic collision between a projectile particle of mass \(m_{\mathrm{A}}\) and a target particle (at rest) of mass \(m_{\mathrm{B}}\), show that the scattering angle, \(\theta_{\mathrm{A}}^{\prime},\) of the projectile \((a)\) can take any value, 0 to \(180^{\circ},\) for \(m_{\mathrm{A}}m_{\mathrm{B}}\).

Short Answer

Expert verified
For \(m_A < m_B\), \(\theta'_A\) can be any angle. For \(m_A > m_B\), \(\theta'_A\) is limited by \(\cos^2\phi = 1-\left(\frac{m_B}{m_A}\right)^2\).

Step by step solution

01

Understand the problem

We need to solve an elastic collision problem where a projectile with mass \(m_A\) collides with a target mass \(m_B\) initially at rest. The goal is to determine how the scattering angle \(\theta'_A\) of the projectile can vary based on the masses of the particles involved.
02

Recall Elastic Collision Principles

In an elastic collision, both momentum and kinetic energy are conserved. Key equations are:1. \( m_A v_{A0} = m_A v_{A}' \cos(\theta'_A) + m_B v_{B}' \cos(\theta'_B) \) (Conservation of Momentum)2. \( \frac{1}{2} m_A v_{A0}^2 = \frac{1}{2} m_A v_{A}'^2 + \frac{1}{2} m_B v_{B}'^2 \) (Conservation of Kinetic Energy)\where \(v_{A0}\) is the initial velocity of particle A, and \(v_A'\), \(v_B'\) are velocities after the collision.
03

Derive the Expression for Scattering Angle (Case m_A < m_B)

For \(m_A < m_B\), the conservation laws allow \(\theta'_A\) to vary from \(0\) to \(180^{\circ}\) because there is no mass restriction preventing these values. Geometrically, the lighter particle can scatter in all directions.
04

Derive the Expression for Scattering Angle (Case m_A > m_B)

For \(m_A > m_B\), however, the analysis changes. Since the projectile is heavier, a critical maximum angle \(\phi\) arises. To find this angle, manipulate the conservation equations to connect \(\cos(\theta'_A)\) to mass ratios:
05

Establish Relationship for Maximum Angle \(\phi\)

When deriving for \(m_A > m_B\), you find:\[ \cos^2(\phi) = 1 - \left(\frac{m_B}{m_A}\right)^2 \]This outcome limits the range of potential scattering angles, meaning \(\theta'_A\) cannot exceed this derived \(\phi\).
06

Conclusion and Final Check

Verify that the math checks out by looping back to verify each scenario. For \(m_A < m_B\), no restrictions enforceable over \(\theta'_A\). For \(m_A > m_B\), the physical conditions are consistent with the existence of a maximum scattering angle \(\phi\), satisfying the derived expression.

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Key Concepts

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

Scattering Angle
In an elastic collision, the scattering angle is the angle at which a projectile changes direction after hitting a target. Importantly, this angle, denoted as \( \theta_A' \), is pivotal in understanding how the directions of the particles change after impact.
  • For massive projectiles ( \( m_A > m_B \)), the scattering is limited to a range defined by critical angles.
  • Conversely, when the target is heavier ( \( m_A < m_B \)), the projectile can scatter freely in a full range from \( 0 \) to \( 180^{\circ} \).
This behavior is due to how mammoth objects can only be deflected to a certain degree by lighter objects. Understanding the scattering angle is critical in analyzing elastic collisions as it helps predict motion paths after collisions.
Momentum Conservation
In the realm of physics, momentum conservation is crucial, especially in studying elastic collisions. Here, momentum is the quantity of motion an object possesses, defined as the product of its mass and velocity.Elastic collision conservation of momentum can be expressed as:- Before collision: \[ m_A v_{A0} = m_A v_{A}' \cos(\theta'_A) + m_B v_{B}' \cos(\theta'_B) \]- After collision: Momentum is redistributed but the total remains constant.This concept means that the overall momentum before and after the collision remains unchanged. This principle allows us to establish the direction and magnitude of velocities post-collision, contributing to understanding scattering angles.
Kinetic Energy Conservation
In elastic collisions, kinetic energy conservation is as vital as momentum conservation. It states that the total kinetic energy of the system remains constant before and after the collision.The kinetic energy conservation equation is:- Before and after collision: \[ \frac{1}{2} m_A v_{A0}^2 = \frac{1}{2} m_A v_{A}'^2 + \frac{1}{2} m_B v_{B}'^2 \]In simple terms, the sum of kinetic energy in both the projectile and target remains unaltered throughout the process.
  • It establishes that energy exchanges strictly occur between particles.
  • This property is fundamental to ensuring no energy is lost, allowing accurate calculations of scattering angles post-collision.
Projectile Mass
The mass of the projectile, denoted as \( m_A \), is a key variable affecting the scattering angle in collisions.
  • When the projectile mass \( m_A \) is less than that of the target \( m_B \), it can move in a wider range, effectively from \( 0 \) to \( 180^{\circ} \).
  • Conversely, if the projectile mass \( m_A \) exceeds the target mass \( m_B \), the scattering range is limited by a maximum angle \( \phi \), determined by the expression:
  • \[ \cos^2(\phi) = 1 - \left(\frac{m_B}{m_A}\right)^2 \]
In essence, the ratio of masses dictates the potential scope of the projectile's scattering! This relationship between projectile and target mass is fundamental for understanding particle collisions in physics.

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Most popular questions from this chapter

You have been hired as an expert witness in a court case involving an automobile accident. The accident involved car A of mass \(1500 \mathrm{~kg}\) which crashed into stationary car \(\mathrm{B}\) of mass \(1100 \mathrm{~kg} .\) The driver of car \(\mathrm{A}\) applied his brakes \(15 \mathrm{~m}\) before he skidded and crashed into car \(\mathrm{B}\). After the collision, car A slid \(18 \mathrm{~m}\) while car \(\mathrm{B}\) slid \(30 \mathrm{~m}\). The coefficient of kinetic friction between the locked wheels and the road was measured to be \(0.60 .\) Show that the driver of car A was exceeding the \(55-\mathrm{mi} / \mathrm{h}(90 \mathrm{~km} / \mathrm{h})\) speed limit before applying the brakes.

(II) A 920 -kg sports car collides into the rear end of a \(2300-\mathrm{kg}\) SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward \(2.8 \mathrm{~m}\) before stopping. The police officer, estimating the coefficient of kinetic friction between tires and road to be \(0.80,\) calculates the speed of the sports car at impact. What was that speed?

(II) A 0.145-kg baseball pitched horizontally at \(32.0 \mathrm{~m} / \mathrm{s}\) strikes a bat and is popped straight up to a height of \(36.5 \mathrm{~m}\). If the contact time between bat and ball is \(2.5 \mathrm{~ms},\) calculate the average force between the ball and bat during contact.

The gravitational slingshot effect. Figure \(9-55\) shows the planet Saturn moving in the negative \(x\) direction at its orbital speed (with respect to the Sun) of \(9.6 \mathrm{~km} / \mathrm{s}\). The mass of Saturn is \(5.69 \times 10^{26} \mathrm{~kg} .\) A spacecraft with mass \(825 \mathrm{~kg}\) approaches Saturn. When far from Saturn, it moves in the \(+x\) direction at \(10.4 \mathrm{~km} / \mathrm{s}\). The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn's gravitational pull.

(II) A child in a boat throws a 5.70 -kg package out horizontally with a speed of \(10.0 \mathrm{~m} / \mathrm{s}\), Fig. 9-37. Calculate the velocity of the boat immediately after, assuming it was initially at rest. The mass of the child is \(24.0 \mathrm{~kg}\) and that of the boat is \(35.0 \mathrm{~kg}\).
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