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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