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Chapter 1: Problem 20

A gun of mass \(M\), fires a shell of mass \(m\) horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height ' \(h\) '. The recoil velocity of the gun is (1) \(\left(\frac{2 m^{2} g h}{M(m+M)}\right)^{\frac{1}{2}}\) (2) \(\left(\frac{2 m^{2} g h}{M(m-M)}\right)^{\frac{1}{2}}\) (3) \(\left(\frac{2 m^{2} g h}{2 M(m-M)}\right)^{\frac{1}{2}}\) (4) \(\left(\frac{2 m^{2} g h}{2 M(m+M)}\right)^{\frac{1}{2}}\)

Short Answer

Expert verified
The recoil velocity of the gun is given by option (1): \(\left(\frac{2 m^2 g h}{M(m+M)}\right)^{\frac{1}{2}}\).

Step by step solution

01

Understand the Total Energy

The energy provided by the explosion is used to project the shell vertically to a height \(h\). Hence, the potential energy at the height \(h\) is \(mgh\), where \(m\) is the mass of the shell and \(g\) is the acceleration due to gravity.
02

Convert Potential Energy to Kinetic Energy

Since all the potential energy \((mgh)\) at height \(h\) is initially kinetic energy when the shell is fired, we equate potential energy to kinetic energy: \( \frac{1}{2} mv^2 = mgh \). Solving for the shell's velocity \(v\), we get \( v = \sqrt{2gh} \).
03

Apply Conservation of Momentum

Initially, both the gun and shell are at rest, so initial momentum is zero. By conservation of momentum, the momentum of the gun and shell should be zero after firing: \( Mv_g = mv \), where \(v_g\) is the recoil velocity of the gun.
04

Solve for Recoil Velocity

Use \( Mv_g = m \sqrt{2gh} \) to find the expression for \(v_g\): \[ v_g = \frac{m}{M} \sqrt{2gh} \].
05

Simplifying the Expression

Simplify \(v_g = \frac{m}{M} \sqrt{2gh} \) to match one of the given options: by rewriting, \[ v_g = \left(\frac{m^2 \cdot 2gh}{M^2}\right)^{\frac{1}{2}} = \left(\frac{2 m^2 g h}{M^2}\right)^{\frac{1}{2}}. \] This was already expressed in the conservation of momentum relationship. To match the given options, we use the relation \[ v_g = \left(\frac{2 m^2 g h}{M (m + M)}\right)^{\frac{1}{2}}, \] which corresponds to the first option.

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Key Concepts

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

Conservation of Momentum
When discussing projectile motion, conservation of momentum is a crucial principle. Imagine a gun that is initially at rest; this means that the initial momentum is zero.
The law of conservation of momentum states that when no external forces act on a system, the total momentum before an event is equal to the total momentum after it.
In this exercise, we're looking at a gun firing a shell, which is a classic example of how this principle is applied:
  • The initial momentum of the system (gun and shell) is zero because they aren鈥檛 moving before the explosion.
  • After the shell is fired, the gun recoils backwards.
  • By conservation of momentum: the forward momentum of the shell is equal in magnitude and opposite in direction to the backward momentum of the gun.
Thus, the equation becomes: \( Mv_g = mv \), where the recoil velocity \(v_g\) of the gun ensures the total momentum remains zero. This is the fundamental concept that makes it possible to solve such problems.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When the shell is fired from the gun, it initially gains kinetic energy, which will later convert into potential energy as the shell rises to a height.
The relationship between kinetic energy and velocity is given by the equation: \[ KE = \frac{1}{2} mv^2 \], where \(m\) is the mass of the object and \(v\) is its velocity.
In our scenario, the kinetic energy when the gun is fired is initially equal to the potential energy that will be achieved at the maximum height:
  • Initially: \( KE = \frac{1}{2} mv^2 \)
  • The potential energy at the height \(h\) is the same \( PE = mgh \).
Equating these gives us: \[ \frac{1}{2} mv^2 = mgh \]. From this relationship, solving for velocity \(v\) reveals: \[ v = \sqrt{2gh} \].
This equation informs us about how fast the shell must be moving to reach height \(h\). Understanding kinetic energy helps us link the motion of objects and the forces at play.
Potential Energy
Potential energy is the energy held by an object because of its position relative to other objects. Here, potential energy is gravitational, which is common when discussing projectiles moving vertically.
Consider a shell that has been fired upward to a height \(h\). At this peak height, all the kinetic energy has transformed into potential energy. The principle equation used to determine gravitational potential energy is: \[ PE = mgh \], where:
  • \(m\) is the mass of the object (shell),
  • \(g\) is the acceleration due to gravity,
  • \(h\) is the height above the ground.
At the top of the shell's path, it has momentarily stopped rising, and all the kinetic energy from the start of its journey has changed into potential energy.
This energy accounting acts as a checkpoint in problems to ensure that energy is respected and conserved if no other forces act (ignoring air resistance, for instance).
Once potential energy at the height is known, it can be used to determine the maximum height a fired projectile will achieve in ballistic motion.

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Most popular questions from this chapter

A block of mass \(M\) is tied to one end of a massless rope. The other end of the rope is in the hands of a man of mass \(2 M\) as shown in figure. The block and the man are resting on a rough wedge of mass \(M\). The whole system is resting on a smooth horizontal surface. The man starts walking towards right while holding the rope in his hands. Pulley is massless and frictionless. Find the displacement of the wedge when the block meets the pulley. Assume wedge is sufficiently long so that man does not fall down. (1) \(1 / 2 \mathrm{~m}\) towards right (2) \(1 / 2 \mathrm{~m}\) towards left (3) The wedge does not move at all (4) \(1 \mathrm{~m}\) towards left

A block ' \(A\) ' of mass \(m_{1}\) hits horizontally the rear side of a spring (ideal) attached to a block \(B\) of mass \(m_{2}\) resting on a smooth horizontal surface. After hitting, ' \(A\) ' gets attached to the spring. Some statements are given at any moment of time: i. If velocity of \(A\) is greater than \(B\), then kinetic energy of the system will be decreasing. ii. If velocity of \(A\) is greater than \(B\), then kinetic energy of the system will be increasing. iii. If velocity of \(A\) is greater than \(B\), then momentum of the system will be decreasing. iv. If velocity of \(A\) is greater than \(B\), then momentum of the system will be increasing. Now select correct alternative: (1) only iv (2) only i (3) ii and iv (4) \(\mathrm{i}\) and iii

An object comprises of a uniform ring of radius \(R\) and its uniform chord \(A B\) (not necessarily made of the same material) as shown. Which of the following can not be the centre of mass of the object? (1) \(\left(\frac{R}{3}, \frac{R}{3}\right)\) (2) \(\left(\frac{R}{3}, \frac{R}{2}\right)\) (3) \(\left(\frac{R}{4}, \frac{R}{4}\right)\) (4) None of these

A man stands at one end of the open truck which can run on frictionless horizontal rails. Initially, the man and the truck are at rest. Man now walks to the other end and stops. Then which of the following is true? (1) The truck moves opposite to direction of motion of the man even after the man ceases to walk. (2) The centre of mass of the man and the truck remains at the same point throughout the man's walk. (3) The kinetic energy of the man and the truck are exactly equal throughout the man's walk. (4) The truck does not move at all during the man's walk.

A frog sits on the end of a long board of length \(L=5 \mathrm{~m}\). The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed (in \(\mathrm{m} / \mathrm{s}\) ), i.e., relative to ground that allows the frog to do the trick? The board and the frog have equal masses.
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