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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 \(A\)

Short Answer

Expert verified
3: The particles move in mutually perpendicular directions after collision.

Step by step solution

01

Understand Elastic Collision

In an elastic collision, both kinetic energy and momentum are conserved. Given that the particles have equal mass, this simplifies the analysis of their final velocities after collision. We denote the initial velocity of particle \(A\) as \(v_{A0}\) and the velocity of particle \(B\) as \(0\) (since it is initially at rest).
02

Analyze Conservation of Momentum

Conservation of momentum along the line of the initial velocity gives us the equation \( mv_{A0} = mv_{A1} \cos(\theta) + mv_{B1} \cos(\phi) \), where \(v_{A1}\) and \(v_{B1}\) are the final velocities of \(A\) and \(B\), and \(\theta\) and \(\phi\) are the angles their paths make with the original direction.
03

Analyze Conservation of Kinetic Energy

The conservation of kinetic energy is expressed as \( \frac{1}{2}mv_{A0}^2 = \frac{1}{2}mv_{A1}^2 + \frac{1}{2}mv_{B1}^2 \). This equation ensures that the total kinetic energy before and after the collision remains the same.
04

Simplify with Conditions of Equal Mass

With equal masses and the equations from the conservation laws, one viable solution is if \(v_{A1} = v_{B1}\) and they move perpendicular to each other. This corresponds to condition (3). Any attempt for both to maintain their direction or for only one to remain stationary violates either momentum or energy conservation.

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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 quantity in physics that describes an object's motion. It is a product of an object's mass and velocity, represented mathematically as \( p = mv \). In any closed system, like our collision scenario, the total momentum before the collision is equal to the total momentum after.In the original problem, particle \(A\) has an initial velocity and thus an initial momentum, while particle \(B\) is at rest, meaning its initial momentum is zero. The conservation law gives us a way to relate the final movements of these particles to their initial conditions:
  • Initial momentum: \( mv_{A0} = mv_{A1} \cos(\theta) + mv_{B1} \cos(\phi) \)
Here, \(v_{A1}\) and \(v_{B1}\) are the velocities after collision, and \(\theta, \phi\) are the angles of their movements relative to the initial direction of \(A\).The importance of this law is that it provides a way to predict the final state of the system without knowing the detailed interactions during the collision. In short, momentum tells us that the combined carrying capacity of motion remains consistent despite individual changes.
Conservation of Kinetic Energy
Kinetic energy is the energy an object possesses because of its motion, calculated using the formula \( KE = \frac{1}{2} mv^2 \). During an elastic collision, kinetic energy is conserved, meaning the total energy before and after the collision remains constant.For our two particles, this conservation is represented as:
  • Initial kinetic energy: \( \frac{1}{2}mv_{A0}^2 \)
  • Final kinetic energy: \( \frac{1}{2}mv_{A1}^2 + \frac{1}{2}mv_{B1}^2 \)
This equation ensures that as energy moves from one particle to another, the sum doesn't change. In the solution to the exercise, we see this conservation principle paired with equal masses leading to them either swapping speeds or moving perpendicular. The beauty of elastic collisions is that they are predictable, allowing us to use energy conservation laws to determine post-collision characteristics as long as initial conditions are well understood.
Perpendicular Motion After Collision
In elastic collisions, when two objects collide and move at right angles, it’s often due to their equal masses, and it's known as perpendicular or orthogonal motion. This unique relationship arises from the conservation laws dictating how the post-collision velocities must sum to maintain the system's initial momentum and energy.After the collision, if particles \(A\) and \(B\) move in directions that form a 90-degree angle with each other, this aligns directly with our momentum and kinetic energy equations discussed earlier. For instance, using trigonometric identities:
  • \( \cos^2(\theta) + \cos^2(\phi) = 1 \), when the angle between their paths is \(90\) degrees, simplifying energy conservation relations.
This path then satisfies a condition well-known in physics where equal masses undergoing elastic impact resolve to right-angle paths, a direct outcome of the conservation laws' constraints. It illustrates how nature finds elegant solutions to complex scenarios and offers an insight into the underlying symmetry of physical laws in action.

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Most popular questions from this chapter

A particle loses \(25 \%\) of its energy during collision with another identical particle at rest. the coefficient of restitution will be (1) \(0.25\) (2) \(\sqrt{2}\) (3) \(\frac{1}{\sqrt{2}}\) (4) \(0.5\)

A particle of mass \(m\) comes down on a smooth inclined plane from point \(B\) at a height of \(h\) from rest. The magnitude of change in momentum of the particle between position \(A\) (just before arriving on horizontal surface) and \(C\) (assuming the angle of inclination of the plane as \(\theta\) with respect to the horizontal) is (1) 0 (2) \(2 m \sqrt{(2 g h)} \sin \theta\) (3) \(2 m \sqrt{(2 g h)} \sin \left(\frac{\theta}{2}\right)\) (4) \(2 m \sqrt{(2 g h)}\)

A ball of mass \(m\) is released from rest relative to elevator at a height \(h_{1}\) above the floor of the elevator. After making collision with the floor of the elevator it rebounces to height h. The coefficient of rectithtion \(h_{2}\). The coefficient of restitution for collision is \(e\). For this situation, mark the correct statement(s). (1) If elevator is moving down with constant velocity \(v_{0}\), then \(h_{2}=e^{2} h_{1}\) (2) If elevator is moving down with constant velocity \(v_{0}\), then \(h_{2}=e^{2} h_{1}-\frac{v_{0}^{2}}{2 g}\) (3) If elevator is moving down with constant velocity \(v_{0}\), then impulse imparted by floor of the elevator to the ball is \(m\left(\sqrt{2 g h_{2}}+\sqrt{2 g h_{1}}+2 v_{0}\right)\) in the upward direction. (4) If elevator is moving with constant acceleration of \(g / 4\) in upward direction, then it is not possible to determine a relation between \(h_{1}\) and \(h_{2}\) from the given information.

A \(5 \mathrm{~kg}\) sphere is connected to a fixed point \(O\) by an inextensible string of length \(5 \mathrm{~m}\). The sphere is resting on a horizontal surface at a distance \(4 \mathrm{~m}\) from \(O\). Sphere is given a vertical velocity \(v_{0}\), and it moves freely till it reaches point \(P\), when string becomes taut. Find the maximum allowable velocity \(v_{0}\) (in \(\mathrm{m} / \mathrm{s}\) ), if impulse of tension in string is not to exceed \(6 \mathrm{~N} \mathrm{~s}\)

Two identical billiard balls undergo an oblique elastic collision. Initially, one of the balls is stationary. If the initially stationary ball after collision moves in a direction which makes an angle of \(37^{\circ}\) with direction of initial motion of the moving ball, then the angle through which initially moving ball will be deflected is (1) \(37^{\circ}\) (2) \(60^{\circ}\) (3) \(53^{\circ}\) (4) \(>53^{\circ}\)

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