/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 21 A body of mass \(1 \mathrm{~kg}\... [FREE SOLUTION] | 91Ó°ÊÓ

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A body of mass \(1 \mathrm{~kg}\) initially at rest, explodes and breaks into three fragments of masses in the ratio \(1: 1: 3\). The two pieces of equal mass fly off perpendicular to each other with a speed of \(15 \mathrm{~ms}^{-1}\) each. What is the velocity of the heavier fragment? (1) \(10 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(5 \sqrt{3} \mathrm{~ms}^{-1}\) (3) \(10 \sqrt{3} \mathrm{~ms}^{-1}\) (4) \(5 \sqrt{2} \mathrm{~ms}^{-1}\)

Short Answer

Expert verified
The velocity of the heavier fragment is \(5\sqrt{2}\) m/s (option 4).

Step by step solution

01

Determine the Masses of the Fragments

Given the initial mass is 1 kg and the ratios of the masses of the fragments are 1:1:3, we can set the masses as follows. Let each of the smaller fragments have a mass \( m \), and hence the heavier fragment has a mass of \( 3m \). Since the total mass is 1 kg: \[ m + m + 3m = 1 \] \[ 5m = 1 \] Therefore, \( m = \frac{1}{5} \) kg. The masses are \( \frac{1}{5} \) kg, \( \frac{1}{5} \) kg, and \( \frac{3}{5} \) kg.
02

Apply Conservation of Momentum

The initial momentum of the system is zero because the body is at rest. After the explosion, the momentum of the system should remain zero. For the first fragment: Momentum = \( \frac{1}{5} \times 15 \) in the x-direction. For the second fragment: Momentum = \( \frac{1}{5} \times 15 \) in the y-direction. Let the velocity of the third fragment be \( v \) in the opposite direction to balance the momentum. Combined equation: \[ \frac{1}{5} \times 15 \hat{i} + \frac{1}{5} \times 15 \hat{j} = \frac{3}{5} \times v \hat{-k} \]
03

Calculate the Velocity of the Heavier Fragment

The magnitude of the momentum for the third fragment is equal to the resultant momentum from the first and second fragments. Using the Pythagorean theorem to find the resultant momentum: \[ P = \sqrt{\left(\frac{1}{5} \times 15\right)^2 + \left(\frac{1}{5} \times 15\right)^2} = \sqrt{\left(3\right)^2 + \left(3\right)^2} = \sqrt{18} = 3\sqrt{2} \] Equating this to the momentum of the third fragment: \[ 3\sqrt{2} = \frac{3}{5} \times v \] Solving for \( v \): \[ v = 5\sqrt{2} \text{ m/s} \]
04

Conclusion

The velocity of the heavier fragment is \( 5\sqrt{2} \text{ m/s} \). Therefore, the correct option is (4) \( 5\sqrt{2} \text{ m/s} \).

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Key Concepts

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

Explosion
When a body explodes, it breaks into several parts. This phenomenon is an interesting scenario of an energy transformation, particularly from the point of view of physics. During an explosion, a single object disintegrates into multiple fragments with significant force. This transformation involves changes in physical properties such as velocity, direction, and distribution of mass.
In the context of our exercise, a 1 kg body undergoes an explosion, breaking into three different pieces. The fragmentation process leads to an entirely new system where conservation laws play a critical role in understanding the resultant motion of each piece. While energy is not necessarily conserved due to the explosive forces, momentum is always conserved in such processes, a fact we can rely on to calculate unknown quantities.
The explosion is characterized by the release of potential energy, which gets converted into the kinetic energy of the fragments, propelling them in different directions. This results in various velocities and sometimes considerably complex dynamics, which can be unravelled using physics principles.
Pythagorean Theorem
The Pythagorean Theorem is a fundamental concept in mathematics. It plays a crucial role in relating the sides of a right-angled triangle. For a triangle with sides of length a, b, and hypotenuse c, the theorem states: \[ a^2 + b^2 = c^2 \]In the exploding body scenario, the pieces of equal mass fly perpendicular to each other. Hence, they form a right-angled triangle with their velocities as the two legs. Using the Pythagorean Theorem, we determine the resultant momentum direction and magnitude.
This application is essential because both fragments shoot out at right angles with the same speed, creating a 90-degree junction. By calculating the square of each momentum component and summing them, the theorem allows us to find the resultant momentum, which dictates the motion necessary to maintain overall momentum balance for the system.
Vector Components
Vectors are quantities that have both magnitude and direction, crucial in understanding motion, especially in multiple dimensions. In our problem, velocity is a vector quantity, making its analysis essential in determining the outcome.
Each of the smaller fragments travels with a velocity of 15 m/s in perpendicular directions. We must decompose these into vector components to analyze their motion. The first component is along the x-axis, while the second is along the y-axis. The third, unknown velocity of the heavier fragment acts to balance these components. Through vector analysis, summing these components helps maintain the law of conservation of momentum.
  • Using vector components ensures we consider both direction and magnitude for all motion-related problems.
  • This concept also acts as a foundation for utilizing the Pythagorean Theorem effectively, as vectors form the sides of the involved triangle.
Mass Ratio in Physics
Mass ratio is a crucial element in understanding how different parts of a system interact during separation events, like explosions. In this problem, a mass ratio of 1:1:3 involves understanding how this proportion affects the momentum distribution.
First, it helps determine the individual masses of the fragmented pieces. Given that the total mass is 1 kg, their masses can be calculated in terms of this ratio, leading to masses of \(\frac{1}{5}\) kg, \(\frac{1}{5}\) kg, and \(\frac{3}{5}\) kg. This calculation allows us to apply the conservation of momentum accurately, as each fragment's mass determines the extent of influence each has on the system's total momentum.
Such ratios are prominent in a variety of physics problems. They are often used to simplify complex interactions by focusing on proportional relationships rather than absolute values, making calculations more straightforward and comprehensible.

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Most popular questions from this chapter

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 blocks of masses \(5 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively, (1) \(4 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (2) \(10 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (3) \(4 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\) (4) \(10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\)

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.

A strip of wood of mass \(M\) and length \(l\) is placed on a smooth horizontal surface. An insect of mass \(m\) starts at one end of the strip and walks to the other end in time \(t\), moving with a constant speed. The speed of the insect as seen from the ground is (1) \(\frac{l}{t}\left(\frac{M}{M+m}\right)\) (2) \(\frac{l}{t}\left(\frac{m}{M+m}\right)\) (3) \(\frac{l}{t}\left(\frac{M}{m}\right)\) (4) \(\frac{l}{t}\left(\frac{m}{M}\right)\)

A man is standing on a plank which is placed on smooth horizontal surface. There is sufficient friction between feet of man and plank. Now man starts running over plank, correct statement is/are (1) Work done by friction on man with respect to ground is negative. (2) Work done by friction on man with respect to ground is positive (3) Work done by friction on plank with respect to ground is positive. (4) Work done by friction on man with respect to plank is zero.

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