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Chapter 1: Problem 17

A vessel at rest explodes breaking it into three pieces Two pieces having equal mass fly off perpendicular to on another with the same speed of \(30 \mathrm{~m} / \mathrm{s}\). The third piece ha three times the mass of each of the other two pieces. Wha is the direction (w.r.t. the pieces having equal masses) an magnitude of its velocity immediately after the explosion? (1) \(10 \sqrt{2}, 135^{\circ}\) (2) \(10 \sqrt{2}, 90^{\circ}\) (3) \(10 \sqrt{2}, 60^{\circ}\) (4) \(10 \sqrt{2}, 30^{\circ}\)

Short Answer

Expert verified
The velocity of the third piece is \( 10\sqrt{2} \) m/s at \( 135^{\circ} \).

Step by step solution

01

Understand conservation of momentum

The total momentum before the explosion is zero (since the vessel is at rest), and by conservation of momentum, the total momentum after the explosion is also zero. We set up the equations to account for this in both x and y directions.
02

Analyze momentum in x and y directions

Let the mass of each small piece be \( m \) and the mass of the third piece be \( 3m \). The speed of the two smaller pieces is \( 30 \:\text{m/s} \), and they travel perpendicular to each other. Thus, the momentum in the x-direction is \( mv_1 - 3mv_3 \cos \theta = 0 \) and in the y-direction \( mv_2 - 3mv_3 \sin \theta = 0 \). Both pieces have velocity magnitude \( v_1 = 30 \) and \( v_2 = 30 \) due to being equal masses going perpendicular.
03

Set up equations for momentum conservation

The equations for momentum conservation become: 1. In the x-direction: \( m \times 30 - 3m \times v_3 \cos \theta = 0 \)2. In the y-direction: \( m \times 30 - 3m \times v_3 \sin \theta = 0 \)We simplify these as follows:1. \( 30 - 3v_3 \cos \theta = 0 \)2. \( 30 - 3v_3 \sin \theta = 0 \)
04

Solve for \( v_3 \)

From the equation \( 30 - 3v_3 \cos \theta = 0 \), we have \( v_3 \cos \theta = 10 \). From the equation \( 30 - 3v_3 \sin \theta = 0 \), we have \( v_3 \sin \theta = 10 \).Using these in Pythagorean Theorem, we see \[ v_3 = \sqrt{(10)^2 + (10)^2} = 10\sqrt{2} \].
05

Determine direction \( \theta \)

The tan ratio for the conservation gives \( \tan \theta = \frac{10}{10} = 1 \). This results in the angle \( \theta = 45^{\circ} \).Since one small piece goes x-direction and another goes y-direction, the angle for the third piece is \( 90 + \theta = 135^{\circ} \) as the major piece must travel opposite to the resultant of x and y components combined.

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Key Concepts

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

Explosion Mechanics
Explosions are fascinating phenomena where a single object at rest suddenly breaks into multiple fragments, instantly setting them in motion. This behavior is governed by the principle known as conservation of momentum. Before the explosion, an object is at rest, meaning its total momentum is zero. After the explosion, the sum of the momenta of all pieces must still equal zero.
In this exercise, a vessel explodes into three pieces. Two of the fragments have equal mass and move away at the same speed perpendicular to each other. The third fragment, with a greater mass, must account for the balance of momentum as dictated by the conservation principle.
  • Total initial momentum = 0 (object at rest)
  • Total final momentum = 0 (sum of individual momenta)
Understanding these dynamics is crucial as it sets the stage for analyzing individual velocities and directions of motion.
Perpendicular Velocity
When two objects are said to travel in perpendicular directions, they move at right angles (90 degrees) to one another. This scenario is common in this problem, where two fragments after an explosion travel perpendicular to one another.
Imagine two pieces flying away from the initial explosion point: one moves horizontally, the other vertically. Both pieces share the same velocity magnitude of 30 m/s but in different directions. This orthogonal movement is significant because it simplifies the calculation of resultant velocity for the third piece.
  • Two pieces are perpendicular to each other.
  • Equal speeds (magnitude) but different directions.
This perpendicular relationship means that the momenta in x and y directions act independently, allowing us to tackle each direction separately within the mathematical problem.
Momentum in X and Y Directions
Momentum is a vector quantity, meaning it has both magnitude and direction. In this exercise, we need to analyze it separately for the x and y axes.
Let's set up our system. If one piece flies off in the x-direction and another in the y-direction, the third piece must ensure the total momentum in each axis sums to zero. Here's how we explore the components:
  • x-direction: momentum of the first small piece is balanced by the third piece's x-component.
  • y-direction: momentum of the second small piece is countered by the third piece's y-component.
This essentially creates a balancing act, where setting up equations like \( mv_1 = 3mv_3 \cos \theta \) and \( mv_2 = 3mv_3 \sin \theta \) helps us find the magnitude and direction the third piece must take to satisfy the laws of motion.
Angle of Resultant Velocity
The angle of resultant velocity is critical to understand in explosion mechanics. It tells us the direction in which the third piece moves after an explosion.
With the velocity components \( v_3 \cos \theta = 10 \) and \( v_3 \sin \theta = 10 \), we used trigonometric identities to find the net speed of this piece. Employing the Pythagorean Theorem, we determine \( v_3 = \sqrt{(10)^2 + (10)^2} = 10\sqrt{2} \).
Next, calculate the angle using the tangent function, \( \tan \theta = \frac{10}{10} = 1 \), leading to an angle of \( \theta = 45^{\circ} \). Since the third piece must travel directly opposite the initial two fragments' resultant motion, the final angle relative to these is \( 135^{\circ} \).
  • Use trigonometric functions for components.
  • Determine the principal angle for opposite motion rationale.
This complete understanding of the angle allows us to fully predict the third piece's trajectory post-explosion.

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Most popular questions from this chapter

Which of the following is/are correct? (1) If centre of mass of three particles is at rest and it is known that two of them are moving along different lines, then the third particle must also be moving. (2) If centre of mass remains at rest, then net work done by the forces acting on the system must be zero. (3) If centre of mass remains at rest, then the net external force must be zero. (4) If speed of centre of mass is changing, then there must be some net work being done on the system from outside.

In a system of particles \(8 \mathrm{~kg}\) mass is subjected to a force of \(16 \mathrm{~N}\) along \(+\) ve \(x\)-axis and another \(8 \mathrm{~kg}\) mass is subjected to a force of \(8 \mathrm{~N}\) along \(+\) ve \(y\)-axis. The magnitude of acceleration of centre of mass and the angle made by it with \(x\)-axis are given, respectively, by (1) \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-2}, \theta=45^{\circ}\) (2) \(3 \sqrt{5} \mathrm{~ms}^{-2}, \theta=\tan ^{-2}\left(\frac{2}{3}\right)\) (3) \(\frac{\sqrt{5}}{2} \mathrm{~ms}^{-2}, \theta=\tan ^{-1}\left(\frac{1}{2}\right)\) (4) \(1 \mathrm{~ms}^{-2}, \theta=\tan ^{-2} \sqrt{3}\)

A uniform wire frame \(\mathrm{ABC}\) is in the shape of an equilateral triangle in \(x y\)-plane with centroid at the origin. (1) If \(A B\) is removed, the centre of mass of the remaining figure is in fourth quadrant. (2) If \(A C\) is removed, the centre of mass of the remaining figure is in first quadrant. (3) If \(B C\) is removed, the centre of mass of the remaining figure is on the positive \(Y\)-axis. (4) If \(A C\) is removed, the centre of mass of the remaining figure is in third quadrant.

A trolley is moving horizontally with a velocity of \(v \mathrm{~m} / \mathrm{s}\) w.r.t. earth. A man starts running in the direction of motion of trolley from one end of the trolley with a velocity \(1.5 v\) \(\mathrm{m} / \mathrm{s}\) w.r.t. the trolley. After reaching the opposite end, the man turns back and continues running with a velocity of \(1.5\) \(v \mathrm{~m} / \mathrm{s}\) w.r.t. trolley in the backward direction. If the length of the trolley is \(L\), then the displacement of the man with respect to earth, measured as a function of time, will attain a maximum value of (1) \(\frac{4}{3} L\) (2) \(\frac{2}{3} L\) (3) \(\frac{5 L}{3}\) (4) \(1.5 L\)

Two planks \(A\) and \(B\) are sliding in the same direction on smooth horizontal surface such that \(A\) is faster than \(B\). Packets of equal weight are transferred from \(A\) to \(B\). Due to this: (1) \(A\) will be accelerated and \(B\) will be retarded (2) \(B\) will be accelerated and \(A\) will be retarded (3) There will be no change in \(A\) but \(B\) will be accelerated (4) There will be no change in \(B\) but \(A\) will be accelerated
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