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After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

Short Answer

Expert verified
The angle between the initial velocities is 90 degrees.

Step by step solution

01

Understand the problem

In this problem, two objects of equal mass collide completely inelastically, meaning they stick together after the collision. They each have the same initial speed but different initial directions, and we want to determine the angle between their initial velocities.
02

Write the momentum conservation equation

The total momentum before the collision must equal the total momentum after the collision. Let mass be \( m \) and initial speed be \( v \). After the collision, the velocity of the combined mass is \( \frac{v}{2} \). The initial velocities of the objects form an angle \( \theta \). The momentum conservation equation is:\[ m\mathbf{v}_1 + m\mathbf{v}_2 = (2m)\frac{v}{2} \]
03

Express initial velocities

Let the initial velocity of object 1 be \( v \) in the direction of the x-axis, so its momentum is \( mv \). Let object 2 have initial velocity \( v \) at an angle \( \theta \) from the x-axis, giving momentum components \( mv\cos(\theta) \) in x and \( mv\sin(\theta) \) in y.
04

Solve for the x-component of momentum

The x-component of momentum is:\[ mv + mv \cos(\theta) = 2m \cdot \frac{v}{2} \]Solve for \( \cos(\theta) \):\[ 1 + \cos(\theta) = 1 \]\[ \cos(\theta) = 0 \]
05

Solve for the y-component of momentum

The y-component of velocity must be zero as there is no vertical velocity after collision:\[ mv\sin(\theta) = 0 \]Since \( v eq 0 \), this implies \( \sin(\theta) = 0 \), consistent with the solution found in the x-component.
06

Calculate the angle

Since \( \cos(\theta) = 0 \), \( \theta = 90^\circ \). The sine is also zero for other angles, but combined with the angles such as \( 0^\circ \) and \( 180^\circ \) that won't satisfy \( \cos(\theta) \). Therefore, \( \theta = 90^\circ \) is appropriate for this motion such that they move to a halt initially at x before moving away with equal contribution along both axes.

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Key Concepts

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

Momentum Conservation
In an inelastic collision, the concept of momentum conservation is essential. Momentum, which is the product of an object's mass and its velocity, remains the same before and after the collision in the absence of external forces. In the given exercise, two objects of identical masses are involved in such a collision. Each object starts with the same speed, but their initial directions are unspecified. Understanding momentum conservation helps us derive the post-collision behavior of these objects.

In mathematical terms, we express momentum before and after the collision using the equation:
  • Before collision: Total initial momentum = \( m\mathbf{v}_1 + m\mathbf{v}_2 \)
  • After collision: Total final momentum = \( (2m)\frac{v}{2} \)
This simplifies the problem to finding the relationship between initial directions that lead to a specific final state. By ensuring the total momentum before and after the collision is equal, we can infer crucial details about the collision, such as the angle between the objects' velocities.
Velocity Components
Velocity components are pivotal when analyzing collisions, especially when objects move in different directions. Each object's velocity can be decomposed into two perpendicular components, usually along the x and y axes. This decomposition allows us to break down complex vector problems into simpler scalar parts.

For instance, if object 1 travels along the x-axis, its momentum is straightforward (\(mv\)). However, if object 2 travels at an angle \(\theta\) from this axis, its velocity is decomposed into:
  • x-component: \(v\cos(\theta)\)
  • y-component: \(v\sin(\theta)\)
Understanding these components allows us to tackle the problem piece by piece. For example, by examining x and y separately, the x-components help solve for cosine, while the y-components ensure the net vertical momentum is zero, confirming no vertical movement post-collision.
Collision Angles
In the context of collisions, angles are significant in determining the outcome of the event. The angle between objects' initial velocities dictate how the combined mass behaves after impact. In our exercise, finding this angle was achieved by using the concepts of trigonometry and momentum conservation.

The insight that \(\cos(\theta) = 0\) arose from examining the x-component conservation equation. This directly implies \(\theta\) is \(90^\circ\), aligning with the unique properties of the cosine function:
  • \(\cos(90^\circ) = 0\)
  • Causes complete inelastic collision where no net horizontal or vertical velocity occurs post-collision
This angle determination uses the essential setup of having equal momentum contributions along both axes post-collision. Such calculations are key in understanding how initial relative motion translates into resultant velocities.

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Most popular questions from this chapter

Figure 9-44 shows an arrangement with an air track, in which a cart is connected by a cord to a hanging block. The cart has mass \(m_{1}=0.600 \mathrm{~kg}\), and its center is initially at \(x y\) coordinates \((-0.500\) \(\mathrm{m}, 0 \mathrm{~m}) ;\) the block has mass \(m_{2}=0.400 \mathrm{~kg}\), and its center is initially at \(x y\) coordinates \((0,-0.100 \mathrm{~m})\). The mass of the cord and pulley are negligible. The cart is released from rest, and both cart and block move until the cart hits the pulley. The friction between the cart and the air track and between the pulley and its axle is negligible. (a) In unit-vector notation, what is the acceleration of the center of mass of the cart-block system? (b) What is the velocity of the com as a function of time \(t ?\) (c) Sketch the path taken by the com. (d) If the path is curved, determine whether it bulges upward to the right or downward to the left, and if it is straight, find the angle between it and the \(x\) axis.

A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?

Jumping up before the elevator hits. After the cable }}\( snaps and the safety system fails, an elevator cab free-falls from a height of \)36 \mathrm{~m}\(. During the collision at the bottom of the elevator shaft, a \)90 \mathrm{~kg}\( passenger is stopped in \)5.0 \mathrm{~ms}\(. (Assume that neither the passenger nor the cab rebounds) What are the magnitudes of the (a) impulse and (b) average force on the passenger during the collision? If the passenger were to jump upward with a speed of \)7.0 \mathrm{~m} / \mathrm{s}$ relative to the cab floor just before the cab hits the bottom of the shaft, what are the magnitudes of the (c) impulse and (d) average force (assuming the same stopping time)?

During a lunar mission, it is necessary to increase the speed of a spacecraft by \(2.2 \mathrm{~m} / \mathrm{s}\) when it is moving at \(400 \mathrm{~m} / \mathrm{s}\) relative to the Moon. The speed of the exhaust products from the rocket engine is \(1000 \mathrm{~m} / \mathrm{s}\) relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are \(1200 \mathrm{~kg}\) and \(1800 \mathrm{~kg} ;\) the magnitude of the impulse on each part from the bolts is \(300 \mathrm{~N} \cdot \mathrm{s}\). With what relative speed do the two parts separate because of the detonation?

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