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Chapter 4: Problem 481

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)

Short Answer

Expert verified
The total energy imparted to the two fragments is (D) \(4.8 \mathrm{KJ}\).

Step by step solution

01

Apply the law of conservation of momentum

The law of conservation of momentum states that the initial momentum of a system must be equal to the final momentum of the system. Initially, the bomb is at rest, so its momentum is 0. After the explosion, it breaks into two pieces with the given masses and velocities. Therefore, we can write the equation: \[0 = m_1v_1 + m_2v_2\] where \(m_1\) and \(m_2\) are the masses of the two fragments, and \(v_1\) and \(v_2\) are their respective velocities.
02

Find the velocity of the larger fragment

We are given the mass and velocity of the smaller fragment (\(m_1 = 1.0 \mathrm{~kg}\), \(v_1 = 80 \mathrm{~m} / \mathrm{s}\)) and the mass of the larger fragment (\(m_2 = 2.0 \mathrm{~kg}\)). We can use these values in the equation from Step 1 to find the velocity of the larger fragment, \(v_2\): \[0 = (1.0 \mathrm{~kg})(80 \mathrm{~m} / \mathrm{s}) + (2.0 \mathrm{~kg})(v_2)\] Solving for \(v_2\), we get: \[v_2 = -\frac{1.0 \cdot 80}{2.0} \mathrm{~m} / \mathrm{s} = -40 \mathrm{~m} / \mathrm{s}\] Notice that the velocity of the larger fragment is negative, indicating that the direction of its motion is opposite to that of the smaller fragment.
03

Calculate the kinetic energy of each fragment

Now we will find the kinetic energy of each fragment, which can be calculated using the formula: \[KE = \frac{1}{2}mv^2\] For the smaller fragment: \[KE_1 = \frac{1}{2}(1.0 \mathrm{~kg})(80 \mathrm{~m} / \mathrm{s})^2 = 3200 \mathrm{~J}\] For the larger fragment: \[KE_2 = \frac{1}{2}(2.0 \mathrm{~kg})(-40 \mathrm{~m} / \mathrm{s})^2 = 1600 \mathrm{~J}\]
04

Calculate the total energy imparted to the fragments

Finally, we add the kinetic energy of each fragment to get the total energy imparted to them during the explosion: \[Total \, Energy = KE_1 + KE_2 = 3200 \mathrm{~J} + 1600 \mathrm{~J} = 4800 \mathrm{~J}\] Since we want the answer in kilojoules (KJ), we can convert it: \[4800 \mathrm{~J} = 4.8 \mathrm{~KJ}\] So, the correct answer is (D) \(4.8 \mathrm{KJ}\).

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