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Chapter 9: Problem 62

Two titanium spheres approach each other head-on with the same speed and collide elastically. After the collision, one of the spheres, whose mass is \(300 \mathrm{~g}\), remains at rest. (a) What is the mass of the other sphere? (b) What is the speed of the two-sphere center of mass if the initial speed of each sphere is \(2.00 \mathrm{~m} / \mathrm{s}\) ?

Short Answer

Expert verified
The mass of the other sphere is 150 g. The center of mass speed is \( \frac{2}{3} \) m/s.

Step by step solution

01

Understand the problem

We need to determine the mass of the second sphere after an elastic head-on collision in which one sphere comes to rest, and then find the speed of the center of mass.
02

Analyze the given data

We know the mass of one sphere is 300 g, which remains at rest after the collision. The initial speed of each sphere is 2.00 m/s.
03

Apply conservation of momentum

For elastic collisions, the total momentum before and after the collision must be the same. Let the mass of the second sphere be \( m_2 \). The initial total momentum is \( m_1v_1 + m_2v_2 \), and the final total momentum will be \( m_1 \times 0 + m_2v_2' \). Since the mass of the sphere that comes to rest is 300 g (0.3 kg), \( m_1 = 0.3 \) kg. The velocities just before the collision are both equal to 2 m/s.
04

Set up the equation for momentum conservation

Using the fact that the total momentum before collision is \((m_1v_1 + m_2v_2)\) and after collision is \( m_2v_2' \): \[ 0.3 \times 2 + m_2 \times (-2) = m_2 \times v_2' \] The sphere which stops after the collision has \( v_1' = 0 \) and the other sphere must have its final velocity \( v_2' \) equal to 2 m/s to conserve momentum.
05

Solve for mass of second sphere

From the momentum equation:\[ 0.6 - 2m_2 = 2m_2 \] Combining terms gives: \[ 0.6 = 4m_2 \] Thus, the mass of the second sphere, \( m_2 = 0.15 \) kg or 150 g.
06

Calculate the center of mass speed

The center of mass speed \( v_{cm} \) is determined by using the formula: \[ v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \] Substitute the known values: \[ v_{cm} = \frac{(0.3 \times 2) + (0.15 \times (-2))}{0.3 + 0.15} \] \[ v_{cm} = \frac{0.6 - 0.3}{0.45} = \frac{0.3}{0.45} = \frac{2}{3} \] The speed of the center of mass is \( \frac{2}{3} \) m/s.

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a core principle in physics. It states that the total momentum in a closed system remains constant if no external forces are acting on it.
In the scenario of our titanium spheres, momentum is conserved before and after the collision. Before the collision, both spheres are moving toward each other with a speed of 2 m/s. The total momentum is calculated by multiplying the mass of each sphere with its velocity.
For our exercise, the sphere with mass 0.3 kg has a momentum of \( 0.3 \times 2 = 0.6 \) kg路m/s. If the second sphere has mass \( m_2 \) and a velocity of \(-2\) m/s (moving in the opposite direction), its momentum is \(-2m_2\).
After the collision, the sphere with 0.3kg mass comes to rest, meaning its momentum is 0. And the second sphere must inherit the total initial momentum, resulting in momentum conservation at play: \[ 0.6 - 2m_2 = 2m_2 \]. Solving this lets us find the mass of the second sphere.
Center of Mass
The center of mass of an object or system is a specific point where the weighted relative position of all the mass in the system is considered.
In our exercise, we want to find the center of mass speed of the two titanium spheres. Even though the spheres collide and their velocities change, the overall motion of their center of mass remains consistent, as there are no external forces acting on the system.
To find the speed of the center of mass, we use the formula:
\[ v_{cm} = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} \]
This takes into account the mass and velocity of both spheres before the collision. By substituting the correct values: \( m_1 = 0.3 \) kg, \( m_2 = 0.15 \) kg, \( v_1 = 2 \) m/s, and \( v_2 = -2 \) m/s, we find: \[ v_{cm} = \frac{0.6 - 0.3}{0.3 + 0.15} = \frac{0.3}{0.45} = \frac{2}{3} \] m/s. This shows that even with the collision, the overall center of mass moves at a stable speed of \( \frac{2}{3} \) m/s.
Kinetic Energy
In elastic collisions, kinetic energy is conserved along with momentum. This is a defining trait of such interactions.
Initially, both spheres have kinetic energy derived from both their masses and velocity: \( KE = \frac{1}{2}mv^2 \). The first sphere (0.3 kg) will have kinetic energy \( \frac{1}{2} \times 0.3 \times (2)^2 = 0.6 \) Joules, and the second due to its mass \( m_2 \) and speed will equal \( \frac{1}{2}m_2(-2)^2 \).
After the collision, the sphere that comes to rest has zero kinetic energy, meaning the second sphere retains all initial kinetic energy:
\( KE_{initial} = KE_{final} \). Hence, \( 0.6 + 2m_2 = 2m_2 \), reinforcing that the energy shifts to the moving sphere wholly.

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Most popular questions from this chapter

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