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The centres of the spheres 1,2 and 3 lie on a single straight line. Sphere 1 is moving with an (initial) velocity \(v_{1}\) directed along this line and hits sphere 2 . Sphere 2, acquiring after collision a velocity \(v_{2}\), hits sphere 3 . Both collisions \(a\) absolutely elastic. What must be the mass of sphere 2 (in \(\mathrm{kg}\) for the sphere 3 to acquire maximum velocity (The mass \(m_{1}\) and \(m_{3}\) of spheres 1 and 3 are \(9 \mathrm{~kg}\) and \(1 \mathrm{~kg}\) respectively?

Short Answer

Expert verified
The mass of sphere 2 must be 3 kg for sphere 3 to acquire maximum velocity.

Step by step solution

01

Conservation of Momentum for First Collision

Consider the first collision between sphere 1 and sphere 2. Let the masses of spheres 1 and 2 be \( m_1 = 9 \text{ kg} \) and \( m_2 \) respectively. The initial velocity of sphere 1 is \( v_1 \) and sphere 2 is initially at rest. Use the conservation of momentum:\[ m_1 v_1 = m_1 u_1 + m_2 u_2 \]where \( u_1 \) and \( u_2 \) are the velocities of spheres 1 and 2 post-collision.
02

Conservation of Kinetic Energy for First Collision

The collision is perfectly elastic, so kinetic energy is conserved:\[ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \]From these two equations, we solve for the velocities \( u_1 \) and \( u_2 \) after the first collision.
03

Expression for Velocities after First Collision

Solving the equations from Steps 1 and 2, we get:\[ u_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1 \]\[ u_2 = \frac{2m_1}{m_1 + m_2} v_1 \]Sphere 2 starts moving with \( u_2 \) after the collision.
04

Conservation of Momentum for Second Collision

Now consider the collision between sphere 2 and sphere 3. Sphere 3 has mass \( m_3 = 1 \text{ kg} \) and is initially at rest. The conservation of momentum for this collision is:\[ m_2 u_2 = m_2 v_2 + m_3 v_3 \]where \( v_2 \) and \( v_3 \) are the velocities after the collision.
05

Conservation of Kinetic Energy for Second Collision

The collision is perfectly elastic, so kinetic energy is also conserved:\[ \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_2 v_2^2 + \frac{1}{2} m_3 v_3^2 \]From these equations, solve for \( v_2 \) and \( v_3 \).
06

Calculate Maximum Velocity for Sphere 3

Solve the equations from Step 4 and 5 to find:\[ v_2 = \frac{m_2 - m_3}{m_2 + m_3} u_2 \]\[ v_3 = \frac{2m_2}{m_2 + m_3} u_2 \]The expression for \( v_3 \) is maximized when \( m_2 = \sqrt{m_1 m_3} \).
07

Calculate Optimal Mass for Sphere 2

Given \( m_1 = 9 \text{ kg} \) and \( m_3 = 1 \text{ kg} \), compute:\[ m_2 = \sqrt{9 \times 1} = 3 \text{ kg} \]

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Key Concepts

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

Conservation of Momentum
Momentum is a fundamental concept in physics that describes the motion of bodies. In an elastic collision, the total momentum is conserved. This means that before and after the collision, the momentum of the entire system remains constant.

For our scenario of three spheres, we apply the conservation of momentum in two instances. The first occurs when sphere 1 collides with sphere 2, and the second occurs when sphere 2 collides with sphere 3.

During the first collision, sphere 1 initially moves with a velocity whereas sphere 2 is stationary. The equation for conservation of momentum becomes:
  • Before collision: The total momentum = \( m_1 v_1 \) (since sphere 2 is at rest)
  • After collision: The total momentum = \( m_1 u_1 + m_2 u_2 \)
Here, \( u_1 \) is the velocity of sphere 1 and \( u_2 \) is that of sphere 2 post-collision.

This principle similarly applies to the second collision between sphere 2 and sphere 3. The equations allow us to understand how velocities transfer between the spheres, crucial for determining the maximum velocity that sphere 3 can achieve.
Kinetic Energy Conservation
In elastic collisions, not only is momentum conserved, but kinetic energy is conserved as well. Kinetic energy is the energy an object possesses due to its motion and is given by \( \frac{1}{2} mv^2 \). For perfectly elastic collisions, the total kinetic energy before and after the collision remains the same.

For the first collision between spheres 1 and 2, the kinetic energy equation is written as:
  • Before collision: Total kinetic energy = \( \frac{1}{2} m_1 v_1^2 \)
  • After collision: Total kinetic energy = \( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \)
Solving these equations for each collision helps us determine the post-collision velocities of the spheres.

Similarly, for the second collision between spheres 2 and 3, the conservation of kinetic energy allows us to find the velocities \( v_2 \) and \( v_3 \) after the impact. This ensures no kinetic energy is lost, which is typical of perfectly elastic interactions.
Maximum Velocity
To maximize sphere 3's velocity, we need to consider the optimal mass of sphere 2. When sphere 2 gains velocity from sphere 1 and transfers it to sphere 3, the efficiency of this transfer is related to the masses of the objects involved.

Through physics principles, particularly the formulas derived from conservation laws, it is shown that the optimal mass of sphere 2 is crucial. For sphere 3 to achieve the maximum velocity, the mass \( m_2 \) of sphere 2 is computed using the geometric mean of the masses of spheres 1 and 3:
  • Optimal mass of sphere 2: \( m_2 = \sqrt{m_1 m_3} \)
With given values, \( m_1 = 9 \text{ kg} \) and \( m_3 = 1 \text{ kg} \), it is calculated that \( m_2 = 3 \text{ kg} \). This mass ensures that sphere 3 obtains the highest possible velocity following the interactions.

This insight highlights how intricately connected the masses and velocities are in a system of elastic collisions, ensuring the most effective transfer of energy and momentum through optimal conditions.

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Most popular questions from this chapter

In an elastic collision between two particles (1) the total kinetic energy of the system is always conserved (2) the kinetic energy of the system before collision is equal to the kinetic energy of the system after collision (3) the linear momentum of the system is conserved (4) the mechanical energy of the system before collision is equal to the mechanical energy of the system after collision

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 \(\left(\cos ^{2} \theta+e^{2} \sin ^{2} \theta\right)\)

A body of mass \(2 \mathrm{~kg}\) moving with a velocity \(3 \mathrm{~m} / \mathrm{s}\) collides with a body of mass \(1 \mathrm{~kg}\) moving with a velocity of \(4 \mathrm{~m} / \mathrm{s}\) in opposite direction. If the collision is head on and completely inelastic, then (1) both particles move together with velocity \((2 / 3) \mathrm{m} / \mathrm{s}\) (2) the momentum of system is \(2 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) throughout (3) the momentum of system is \(10 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) (4) the loss of \(\mathrm{KE}\) of system is \((49 / 3) \mathrm{J}\)

A small ball thrown at an initial velocity \(u=25 \mathrm{~m} / \mathrm{s}\) directed at an angle \(\theta=37^{\circ}\) above the horizontal collides elastically with a vertical massive smooth wall moving with a uniform horizontal velocity \(u / 5\) towards the ball. After collision with the wall the ball returns to the point from where it was thrown. Determine the time \(t\) (in \(\mathrm{s}\) ) from the beginning of motion of the ball to the moment of its impact with the wall. (Take \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

A particle \(A\) suffers an oblique elastic collision with a particle \(B\) that is at rest initially. If their masses are the same, then after the collision (1) their KE may be equal (2) \(A\) continues to move in the original direction while \(B\) remains at rest (3) they will move in mutually perpendicular directions (4) \(A\) comes to rest and \(B\) starts moving in the direction of the original motion of \(A\)

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