Question

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 $ms^{-1}$ and the second part of mass 2 kg moves with 8 $ms^{-1}$ speed. If the third part flies off with 4 $ms^{-1}$ speed, then its mass is

• A. 3 kg
• B. 5 kg
• C. 7 kg
• D. 17 kg

Answer 1

The Correct Option is B

To solve this problem, we are going to use the principle of conservation of linear momentum. Let's denote the masses and velocities as follows:

• m_1 = 1 \text{ kg} (mass of the first part)
• v_1 = 12 \text{ ms}^{-1} (velocity of the first part)
• m_2 = 2 \text{ kg} (mass of the second part)
• v_2 = 8 \text{ ms}^{-1} (velocity of the second part)
• m_3 = ? (mass of the third part, which we need to find)
• v_3 = 4 \text{ ms}^{-1} (velocity of the third part)

The parts go off at right angles in a horizontal plane. Therefore, we can treat their momentum components as independent in two perpendicular directions (say, x and y axes).

By the conservation of linear momentum in the x-direction:

• The momentum of the first part is: p_{1x} = m_1 \cdot v_1 = 1 \cdot 12 = 12 \text{ kg ms}^{-1}
• The second part, moving at a right angle, has no momentum in the x-direction: p_{2x} = 0
• The momentum of the third part in x-direction is: p_{3x} = m_3 \cdot v_{3x}

By conservation of momentum in the x-direction:

12 = m_3 \cdot v_{3x}

In the y-direction:

• The momentum of the first part in y-direction is zero because it moves only on the x-axis: p_{1y} = 0
• The momentum of the second part is: p_{2y} = m_2 \cdot v_2 = 2 \cdot 8 = 16 \text{ kg ms}^{-1}
• The momentum of the third part in y-direction is: p_{3y} = m_3 \cdot v_{3y}

By conservation of momentum in the y-direction:

16 = m_3 \cdot v_{3y}

Since v_3 = 4 \text{ ms}^{-1}, we use the Pythagorean theorem to find v_{3x} and v_{3y}:

v_3^2 = v_{3x}^2 + v_{3y}^2

Substituting the momentum equations for balance, we solve for m_3:

\begin{align*} v_{3x} &= \frac{12}{m_3}, \\ v_{3y} &= \frac{16}{m_3}. \end{align*}

(v_3)^2 = (\frac{12}{m_3})^2 + (\frac{16}{m_3})^2

(4)^2 = \frac{144 + 256}{m_3^2}

16 = \frac{400}{m_3^2}

m_3^2 = \frac{400}{16} = 25

m_3 = 5 \text{ kg}

Thus, the mass of the third part is 5 kg.