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Chapter 1: Problem 41

A bomb of mass \(3 m\) is kept inside a closed box of mass \(3 x\) and length \(4 L\) at its centre. It explodes in two parts of mass \(m\) and \(2 m .\) The two parts move in opposite directions and stick to the opposite sides of the walls of box. The box is kept on a smooth horizontal surface. What is the distance moved by the box during this time interval? (1) 0 (2) \(\frac{L}{6}\) (3) \(\frac{L}{12}\) (4) \(\frac{L}{3}\)

Short Answer

Expert verified
The distance moved by the box is 0.

Step by step solution

01

Define the system

The system consists of a bomb of mass \(3m\) inside a box of mass \(3x\) and length \(4L\).
02

Apply conservation of momentum

Since the box is on a smooth surface, it is an isolated system and the total linear momentum is conserved. Before the explosion, the system is at rest, so the total momentum is zero. After the explosion, the mass \(m\) moves in one direction and the \(2m\) mass moves in the opposite direction. Therefore, the momentum of the box should keep the total momentum zero. If \(v_1\) and \(v_2\) are the velocities of masses \(m\) and \(2m\) respectively, and \(V\) is the velocity of the box, the momentum conservation equation is:\[ m \cdot (-v_1) + 2m \cdot v_2 + 3x \cdot V = 0 \] Given that \(v_1 = 2v_2\) (due to the system structure and momentum conservation to the mass ratio), substitute this into the equation.
03

Solve for box velocity \(V\)

Using the relations and assuming \(v_1 = 2v_2\), we have:\[ m \cdot (-2v_2) + 2m \cdot v_2 + 3x \cdot V = 0 \]Simplify:\[ -2mv_2 + 2mv_2 + 3xV = 0 \]Thus,\[ 3xV = 0 \]Therefore, \(V = 0\).
04

Calculate the displacement of the box

Since the velocity of the box \(V = 0\), the box doesn't move. Thus the distance moved by the box is 0.

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Key Concepts

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

Linear Momentum
Linear momentum is a fundamental concept in physics, essential for understanding the motion of objects. It can be simply defined as the product of an object's mass and its velocity. Mathematically, it is expressed as:
  • Linear Momentum = Mass 脳 Velocity
In any given system, the total linear momentum is the vector sum of the momenta of all the objects within the system.
What's crucial to know is that linear momentum is a conserved quantity in physics, which means that in the absence of external forces, the total linear momentum of a system remains constant.
In the exercise at hand, a bomb inside a box explodes into two parts. Before the explosion, the bomb is stationary, and thus the system has a total linear momentum of zero. After the explosion, even though the bomb fragments move, the linear momentum still adds up to zero, which must be accounted for by the motion of the box. This fundamental principle helps us conclude that if no external forces act on the system, the center of mass does not accelerate.
Isolated System
An isolated system is defined as a physical system without any external influence. This means there are no external forces acting on the bodies within the system, allowing for the conservation of total linear momentum.
In our exercise, we assume the box and the bomb inside it form an isolated system. This assumption is crucial, as it means no external forces are acting upon the system, like friction or air resistance, which might otherwise alter the system's momentum.
Because the box is on a smooth horizontal surface, it offers no resistance to movement, supporting the system's isolation.
  • Total momentum before explosion: 0
  • Total momentum after explosion: 0 (assuming the box may move to compensate for the momentum of the bomb fragments)
Recognizing an isolated system allows us to use the conservation laws to solve problems like the one presented in the exercise.
Explosion Dynamics
Explosion dynamics concern the behavior of objects and forces involved during an explosion. Understanding this phenomenon requires insight into how mass and velocity interact following a detonation.
In the problem given, a bomb explodes inside the box into two distinct masses. The mass ratio of the fragments鈥攐ne part being twice as massive as the other鈥攄irectly affects how they move in opposite directions after the explosion. This is maintaining momentum conservation in an isolated system.
Explosions often involve instantaneous release of energy, making them complex. However, in this exercise, we simplify it to involve only momentum conservation without energy considerations.
  • The lighter piece (mass \(m\)) has twice the speed of the heavier piece (mass \(2m\)).
  • This inverse relationship between mass and velocity is critical to ensuring the total momentum before and after the explosion remains zero.
In conclusion, explosion dynamics for isolated systems enable us to predict outcomes like movement or rest states post-explosion, as demonstrated by the box's lack of movement.

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Most popular questions from this chapter

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) \(10 \sqrt{2}, 30^{\circ}\)

An object of mass \(10 \mathrm{~kg}\) is launched from the ground \(t=0\), at an angle of \(37^{\circ}\) above the horizontal with a speed 0 \(30 \mathrm{~m} / \mathrm{s}\). At some time after its launch, an explosion splits the projectile into two pieces. One piece of mass \(4 \mathrm{~kg}\) is observed at \((105 \mathrm{~m}, 43 \mathrm{~m})\) at \(t=2 \mathrm{~s}\). Find the location of second piece at \(t=2 \mathrm{~s}\). (1) \((10,2)\) \((2)(48,16)\) (3) \((10,-2)\) (4) Information insufficient

Figure shows a hollow cube of side ' \(a\) ' and volume ' \(V\). There is a small chamber of volume \(V / 4\) in the cube as shown. The chamber is completely filled by \(m \mathrm{~kg}\) of water. Water leaks through a hole \(H\) and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is (1) \(\frac{1}{2} m g a\) (2) \(\frac{3}{8} m g a\) (3) \(\frac{5}{8} m g a\) (4) \(\frac{1}{8} m g a\)

A circular plate of uniform thickness has a diameter of \(28 \mathrm{~cm}\). A circular portion of diameter \(21 \mathrm{~cm}\) is removed from the plate as shown. \(O\) is the centre of mass of complete plate. The position of centre of mass of remaining portion will shift towards left from 'O' by (1) \(5 \mathrm{~cm}\) (2) \(9 \mathrm{~cm}\) (3) \(4.5 \mathrm{~cm}\) (4) \(5.5 \mathrm{~cm}\)

A ring of mass \(M=90\) gram and radius \(R=3\) meter is kept on a frictionless horizontal surface such that its plane is parallel to horizontal plane. A particle of mass \(m=10\) gram is placed in contact with the inner surface of ring as shown figure. An initial velocity \(v=\sqrt{2} \mathrm{~m} / \mathrm{s}\) is given to the particle along the tangent of the ring. Find the magnitude of the force of interaction (in milli-newton) between the particle and ring.
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