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Chapter 4: Problem 480

A bomb of mass \(10 \mathrm{~kg}\) explodes into 2 pieces of mass \(4 \mathrm{~kg}\) and \(6 \mathrm{~kg}\). The velocity of mass \(4 \mathrm{~kg}\) is \(1.5 \mathrm{~m} / \mathrm{s}\), the K.E. of mass \(6 \mathrm{~kg}\) is ....... (A) \(3.84 \mathrm{~J}\) (B) \(9.6 \mathrm{~J}\) (C) \(3.00 \mathrm{~J}\) (D) \(2.5 \mathrm{~J}\)

Short Answer

Expert verified
The kinetic energy of mass \(6\mathrm{~kg}\) is \(3.00 \mathrm{~J}\).

Step by step solution

01

Apply the Principle of Conservation of Linear Momentum

According to the Principle of Conservation of Linear Momentum, the total momentum before the explosion is equal to the total momentum after the explosion. In this case, the initial momentum of the bomb is zero, so the combined momentum of the two pieces after the explosion must also be zero.
02

Calculate the velocity of mass \(6\mathrm{~kg}\)

Let the velocity of the mass \(6\mathrm{~kg}\) be \(v\). The combined momentum of the two pieces after the explosion can be expressed as: \(4\mathrm{~kg} \cdot 1.5\mathrm{~m/s} = 6\mathrm{~kg} \cdot v\) Now, solve for \(v\): \(v = \dfrac{4\mathrm{~kg} \cdot 1.5\mathrm{~m/s}}{6\mathrm{~kg}}\) \(v = 1\mathrm{~m/s}\) So the velocity of the mass \(6\mathrm{~kg}\) is \(1\mathrm{~m/s}\).
03

Calculate the kinetic energy of mass \(6\mathrm{~kg}\)

Now that we have the velocity of mass \(6\mathrm{~kg}\), we can calculate its kinetic energy using the formula: \(KE = \dfrac{1}{2} m v^2\) where \(KE\) is the kinetic energy, \(m\) is the mass and \(v\) is the velocity. Substitute the given values: \(KE = \dfrac{1}{2} \cdot 6\mathrm{~kg} \cdot (1\mathrm{~m/s})^2\) \(KE = \dfrac{1}{2} \cdot 6\mathrm{~kg} \cdot 1\mathrm{~m^2/s^2}\) \(KE = 3\mathrm{~J}\) The kinetic energy of mass \(6\mathrm{~kg}\) is \(3\mathrm{~J}\).
04

Select the correct option

From our calculations, we have found that the kinetic energy of mass \(6\mathrm{~kg}\) is \(3\mathrm{~J}\). Therefore, the correct option is: (C) \(3.00 \mathrm{~J}\)

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics defined as the product of an object's mass and its velocity. It is a vector quantity, which means it has both magnitude and direction. In our problem, the bomb initially at rest has zero momentum before exploding.

This leads us to the Principle of Conservation of Linear Momentum, which states the total momentum of a closed system remains constant, provided no external forces act upon it. After the explosion, the momentum of the pieces must add up to zero because the system was at rest initially. This principle allows us to solve for unknowns, such as the velocity of one of the pieces.
  • Formula: \[ p = m imes v \] where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
  • In explosions, each piece of debris has its momentum, balancing out to zero when added together.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. It's a scalar quantity and is expressed as one-half the product of an object鈥檚 mass and the square of its velocity.

In our example, once the velocity of a particular mass piece is determined, its kinetic energy can be calculated. Knowing that the conservation of momentum is satisfied in the explosion, we find the kinetic energy to ensure that we're understanding energy distribution in the explosion.
  • Formula: \[ KE = \frac{1}{2} m v^2 \] where \( KE \) is kinetic energy, \( m \) is mass, and \( v \) is velocity.
  • It helps measure how much work an object in motion can accomplish when it collides with something, separated in this problem as energy values for each piece.
Explosion Dynamics
Explosion dynamics is the study of the behavior of materials and energy release in an explosion. It involves understanding how the energy from the explosion is distributed amongst the fragments. In this problem, we see a practical example.

The bomb splits into two pieces, with each piece moving with its velocity and possessing its kinetic energy. Explosion dynamics generally involve high-energy transitions, causing objects to break apart and fly off with significant speed.
  • The total kinetic energy after the explosion is shared between all fragments.
  • Understanding these dynamics helps in different practical fields, from investigating fireworks to forensic science.
  • Key factors include how much energy each piece gets and their resulting speeds.

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Most popular questions from this chapter

A ball is allowed to fall from a height \(20 \mathrm{~m}\). If there is \(30 \%\) loss of energy due to impact, then after one impact ball will go up to (A) \(18 \mathrm{~m}\) (B) \(16 \mathrm{~m}\) (C) \(12 \mathrm{~m}\) (D) \(14 \mathrm{~m}\)

A body falls freely under the action of gravity from a height \(\mathrm{h}\) above the ground. $$ \begin{array}{|l|l|} \hline \text { Column - i } & \text { Column - ii } \\ \hline \text { (a) P.E. }=2 \text { (K.E.) } & \text { (P) constant at every point } \\ \hline \text { (b) P.E. }=\text { K.E. } & \text { (Q) at height }(\mathrm{h} / 3) \\ \hline \text { (c) P.E. }=(1 / 2) \text { (K.E.) } & \text { (R) at height }(2 \mathrm{~h} / 3) \\ \hline \text { (d) P.E.+ K.E. } & \text { (S) at height }(\mathrm{h} / 2) \\ \hline \end{array} $$ (A) \(a-P, b-Q, c-R, d-S\) (B) \(\mathrm{a}-\mathrm{Q}, \mathrm{b}-\mathrm{P}, \mathrm{c}-\mathrm{S}, \mathrm{d}-\mathrm{R}\) (C) \(\mathrm{a}-\mathrm{S}, \mathrm{b}-\mathrm{R}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}\) (D) \(\mathrm{a}-\mathrm{R}, \mathrm{b}-\mathrm{S}, \mathrm{c}-\mathrm{Q}, \mathrm{d}-\mathrm{P}\)

A spring of spring constant \(10^{3} \mathrm{~N} / \mathrm{m}\) is stretched initially \(4 \mathrm{~cm}\) from the unscratched position. How much the work required to stretched it further by another \(5 \mathrm{~cm}\) ? (A) \(6.5 \mathrm{NM}\) (B) \(2.5 \mathrm{NM}\) (C) \(3.25 \mathrm{NM}\) (D) \(6.75 \mathrm{NM}\)

\(1 \mathrm{~kg}\) apple gives \(25 \mathrm{KJ}\) energy to a monkey. How much height he can climb by using this energy if his efficiency is \(40 \%\). (mass of monkey \(=25 \mathrm{~kg}\) and \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\) ) (A) \(20 \mathrm{~m}\) (B) \(4 \mathrm{~m}\) (C) \(30 \mathrm{~m}\) (D) \(40 \mathrm{~m}\)

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) \(10 \mathrm{~m}\)
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