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Magazine Chapter 2: Problem 20
Two particles of equal masses moving with same speed collide perfectly inelastically. After the collision the combined mass moves with half of the speed of the individual masses. The angle between the initial momenta of individual particle is (1) \(60^{\circ}\) (2) \(90^{\circ}\) (3) \(120^{\circ}\) (4) \(45^{\circ}\)
Short Answer
Expert verified
The angle is \( 120^{\circ} \). (Option 3)
Step by step solution
01
Initial momentum expression
Consider two particles, each with mass \( m \) and initial speed \( v \). They move with an angle \( \theta \) between their directions. The initial momentum of the system is \( \vec{p}_1 + \vec{p}_2 = m\vec{v}_1 + m\vec{v}_2 \). To simplify, assume \( \vec{v}_1 \) is along x-axis and \( \vec{v}_2 \) makes an angle \( \theta \) relative to \( \vec{v}_1 \). The components are \( \vec{v}_1 = v\hat{i} \) and \( \vec{v}_2 = v (\cos \theta \hat{i} + \sin \theta \hat{j}) \). Hence, \( \vec{p}_{initial} = m v \hat{i} + m (v \cos \theta \hat{i} + v \sin \theta \hat{j}) = mv(1+\cos \theta)\hat{i} + mv \sin \theta \hat{j} \).
02
Final momentum expression
After the collision, the particles stick together and move with speed \( \frac{v}{2} \) in a certain direction. The combined mass is \( 2m \), so the final momentum is \( \vec{p}_{final} = 2m \cdot \frac{v}{2} \vec{v}_{final} \). This equals \( mv \vec{v}_{final} \), and in terms of direction \( \vec{v}_{final} = \cos \phi \hat{i} + \sin \phi \hat{j} \) gives \( \vec{p}_{final} = mv(\cos \phi \hat{i} + \sin \phi \hat{j}) \).
03
Equating initial and final momenta components
Since momentum is conserved, equate the components from Steps 1 and 2. For x-components, \( mv(1+\cos \theta) = mv \cos \phi \). Simplifying, we get \( 1 + \cos \theta = \cos \phi \). For y-components, \( mv \sin \theta = mv \sin \phi \), which simplifies to \( \sin \theta = \sin \phi \).
04
Analyzing angle \( \phi \) and solving for \( \theta \)
With equations \( 1 + \cos \theta = \cos \phi \) and \( \sin \theta = \sin \phi \), we note that since the final velocity magnitude is \( \frac{v}{2} \), it is a factor of the average speed, so \( \phi \) must be such that \( \cos \phi = \frac{1}{2} \) because the equation \( 1 + \cos \theta = \cos \phi = \frac{1}{2} \) results from equalizing average x-direction residuals. \( \cos \phi = \frac{1}{2} \) occurs at \( \phi = 60^{\circ} \) or \( \phi = 300^{\circ} \), valid overall symmetry-wise. Hence, substituting gives \( 1 + \cos \theta = \frac{1}{2} \), leading to \( \cos \theta = -\frac{1}{2} \). The known angle \( \theta \) for which \( \cos \theta = -\frac{1}{2} \) is \( 120^{\circ} \).
05
Final Step: Conclusion using trigonometry
Thus, the angle \( \theta \) between the initial momenta is \( 120^{\circ} \), corresponding to an inelastic collision resulting in half-speed of the combined mass post collision. Thus, the correct answer is option (3).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Momentum Conservation
In physics, the principle of momentum conservation states that the total momentum of a closed system remains constant if no external forces are acting on it. This principle is particularly useful when analyzing collisions, such as the inelastic collision discussed in the exercise.
In this specific case, two particles collide and stick together, forming a single mass moving at a different velocity. Before the collision, each particle has its own momentum, and after the collision, their combined momentum is shared by the larger mass. This creates an opportunity to apply momentum conservation.
In this specific case, two particles collide and stick together, forming a single mass moving at a different velocity. Before the collision, each particle has its own momentum, and after the collision, their combined momentum is shared by the larger mass. This creates an opportunity to apply momentum conservation.
- Before the collision: Each particle has a momentum of magnitude \(mv\) in possibly different directions.
- After the collision: The combined mass of \(2m\) moves with a unified momentum.
Vector Components
Understanding vector components is crucial when dealing with motions and forces in physics, especially in multi-dimensional problems like this collision question.
Vectors are quantities that have both magnitude and direction. When dealing with vector quantities such as velocity and momentum, it's often helpful to break them down into components along different axes, usually the x and y axes in a 2D framework.
Vectors are quantities that have both magnitude and direction. When dealing with vector quantities such as velocity and momentum, it's often helpful to break them down into components along different axes, usually the x and y axes in a 2D framework.
- For the initial momentum of particle 1, it is simplified along the x-axis so it is \(mv\hat{i}\).
- Particle 2, not along the x-axis, is split into components: \(mv\cos\theta\hat{i}\) along the x-axis and \(mv\sin\theta\hat{j}\) along the y-axis.
Trigonometry in Physics
Trigonometry plays an essential role in physics, especially when dealing with problems involving angles and breakdown of forces. In our exercise, trigonometric identities help solve for angles between the momenta of two particles.
By utilizing basic trigonometric functions such as sine and cosine, we translate the physics of movement and forces into mathematical equations that describe those relationships.
By utilizing basic trigonometric functions such as sine and cosine, we translate the physics of movement and forces into mathematical equations that describe those relationships.
- The equation \(1 + \cos \theta = \cos \phi\) links the angle of the initial momenta with the angle after the collision.
- Using the identity \(\cos \phi = \frac{1}{2}\), commonly known values tell us the angle \(\phi\) corresponds to either \(60^{\circ}\) or \(300^{\circ}\).
