/*! 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 477 A bomb of \(12 \mathrm{~kg}\) di... [FREE SOLUTION] | 91Ó°ÊÓ

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A bomb of \(12 \mathrm{~kg}\) divides in two parts whose ratio of masses is \(1: 4 .\) If kinetic energy of smaller part is \(288 \mathrm{~J}\), then momentum of bigger part in \(\mathrm{kgm} / \mathrm{sec}\) will be (A) 38 (B) 72 (C) 108 (D) Data is incomplete

Short Answer

Expert verified
The momentum of the bigger part is \(72 \mathrm{~kgm}/\mathrm{sec}\).

Step by step solution

01

Find the masses of the two parts

We are given that the bomb with mass 12 kg divides into two parts. We will find the masses of both parts by using the given ratio 1:4. Divide \(12 \mathrm{~kg}\) in the given ratio: For Smaller part, its mass = \(\frac{1}{(1+4)} \times 12 \mathrm{~kg}\) For Bigger part, its mass = \(\frac{4}{(1+4)} \times 12 \mathrm{~kg}\)
02

Calculate the velocity of the smaller part

The kinetic energy of the smaller part is given as 288 J. To find the velocity of the smaller part, we will use the formula for kinetic energy: \(K.E. = \frac{1}{2}mv^2\) where \(K.E.\) is the kinetic energy, \(m\) is the mass, and \(v\) is the velocity. Rearrange the equation to find velocity: \(v = \sqrt{\frac{2K.E.}{m}}\) Plug in the values, we have: \(v = \sqrt{\frac{2(288 J)}{mass ~of~ small ~part}}\)
03

Use conservation of momentum

According to the conservation of momentum, \(momentum ~before ~ explosion = momentum ~after ~ explosion\) Before the explosion, as the bomb is at rest, its momentum is zero. So, after the explosion, the total momentum is also zero. Smaller part momentum + Bigger part momentum = 0 Calculate the smaller part momentum: \(momentum ~of ~smaller ~part = mass ~of ~smaller ~part \times velocity ~of ~smaller ~part\) So we have, \(momentum ~of ~bigger ~part = -momentum ~of ~smaller ~part\)
04

Calculate the momentum of the bigger part

We can now calculate the momentum of the bigger part by plugging in the values from previous steps: \(momentum ~of ~bigger ~part = -momentum ~of ~smaller ~part = -mass ~of ~smaller ~part \times velocity ~of ~smaller ~part\) Plug in the values, and we have the momentum of the bigger part. Finally, compare the result with the given options to select the correct answer.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion. It is calculated using the formula: \( K.E. = \frac{1}{2} m v^2 \), where:
  • \( K.E. \) is the kinetic energy,
  • \( m \) is the mass of the object, and
  • \( v \) is its velocity.
To understand this concept, imagine a moving car. The faster it moves and the heavier it is, the more kinetic energy it has.
In the context of our exercise, the smaller fragment of the bomb has a given kinetic energy of 288 J. To find the velocity, we rearrange the kinetic energy formula to solve for \( v \). The calculation goes as follows: \[ v = \sqrt{\frac{2K.E.}{m}} \] This gives us a way to calculate how fast the smaller piece is moving based on its energy and mass.
Mass Ratio
Mass ratio refers to the relationship between the masses of two or more parts. In the exercise, the bomb splits into two parts with a mass ratio of 1:4. This means the mass of one fragment is one part of a total of five parts, and the other is four parts.
To find the actual masses of the fragments from the total 12 kg, use the following calculation:
  • Mass of the smaller part: \( \frac{1}{(1+4)} \times 12 \text{ kg} \)
  • Mass of the bigger part: \( \frac{4}{(1+4)} \times 12 \text{ kg} \)
This allows you to break down the total mass into the correct proportions based on the given ratio. Understanding mass ratios is crucial in numerous physics problems where distribution of mass affects outcome calculations, such as momentum and inertia.
Momentum
Momentum is a key concept in physics, defined as the product of an object's mass and its velocity. It's represented by the formula: \[ p = m \times v \], where:
  • \( p \) is momentum,
  • \( m \) is mass, and
  • \( v \) is velocity.
Momentum is a vector quantity, meaning it has both magnitude and direction. In our exercise, since the bomb was at rest before the explosion, its total initial momentum was zero. This means the momentum of the bigger part must be equal and opposite to that of the smaller part to maintain the overall system momentum as zero after the explosion. We use this conservation principle to find the momentum of the bigger part by first calculating the momentum of the smaller part and then reversing the direction.
Velocity Calculation
Velocity is an important factor in physics problems when dealing with moving objects. To calculate velocity in such exercises, we often rearrange kinetic or momentum formulas.In our context, after calculating the mass of the smaller fragment and knowing its kinetic energy, we calculate its velocity using: \[ v = \sqrt{\frac{2K.E.}{m}} \].
Velocity directly affects both kinetic energy and momentum, making its accurate calculation crucial. Once you know the velocity of one part of the system, use it to calculate the momentum of that part. As an example in this task, the velocity of the smaller part helps to calculate its momentum. With the principle of conservation of momentum, the bigger part's velocity can indirectly be inferred, which concludes our problem-solving process.

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Most popular questions from this chapter

A spring gun of spring constant \(90 \times 10^{2} \mathrm{~N} / \mathrm{M}\) is compressed \(4 \mathrm{~cm}\) by a ball of mass \(16 \mathrm{~g}\). If the trigger is pulled, calculate the velocity of the ball. (A) \(60 \mathrm{~m} / \mathrm{s}\) (B) \(3 \mathrm{~m} / \mathrm{s}\) (C) \(90 \mathrm{~m} / \mathrm{s}\) (D) \(30 \mathrm{~m} / \mathrm{s}\)

If a man increase his speed by \(2 \mathrm{~m} / \mathrm{s}\), his \(\mathrm{K} . \mathrm{E}\). is doubled, the original speed of the man is (A) \((2+2 \sqrt{2}) \mathrm{m} / \mathrm{s}\) (B) \((2+\sqrt{2}) \mathrm{m} / \mathrm{s}\) (C) \(4 \mathrm{~m} / \mathrm{s}\) (D) \((1+2 \sqrt{2}) \mathrm{m} / \mathrm{s}\)

A ball of mass \(5 \mathrm{~kg}\) is striding on a plane with initial velocity of \(10 \mathrm{~m} / \mathrm{s}\). If co-efficient of friction between surface and ball is \((1 / 2)\), then before stopping it will describe \(\ldots \ldots\) \(\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (A) \(12.5 \mathrm{~m}\) (B) \(5 \mathrm{~m}\) (C) \(7.5 \mathrm{~m}\) (D) \(10 \mathrm{~m}\)

If the K.E. of a body is increased by \(44 \%\), its momentum will increase by....... (A) \(20 \%\) (B) \(22 \%\) (C) \(2 \%\) (D) \(120 \%\)

Four identical balls are lined in a straight grove made on a horizontal frictionless surface as shown. Two similar balls each moving with a velocity v collide elastically with the row of 4 balls from left. What will happen (A) One ball from the right rolls out with a speed \(2 \mathrm{v}\) and the remaining balls will remain at rest. (B) Two balls from the right roll out speed \(\mathrm{v}\) each and the remaining balls will remain stationary. (C) All the four balls in the row will roll out with speed \(\mathrm{v}(\mathrm{v} / 4)\) each and the two colliding balls will come to rest. (D) The colliding balls will come to rest and no ball rolls out from right.

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