Uncategorized: Practice Problem & Solution
A body of mass \( 5 \text{ kg} \) explodes at rest into three fragments with masses in the ratio \( 1 : 1 : 3 \). The fragments with equal masses fly in mutually perpendicular directions with speeds of \( 21 \text{ m/s} \). The velocity of heaviest fragment in \( text{m/s} \) will be: (1989)
Solution Explained:
To solve this problem, we apply the core principles of Uncategorized. Understanding the underlying formula is key to arriving at the correct answer below:
Masses \( m_1 = 1 \text{ kg}, m_2 = 1 \text{ kg}, m_3 = 3 \text{ kg} \). Momentum of first two parts: \( P_1 = 1 \text{ kg} \times 21 \text{ m/s} = 21 \text{ Ns} \), \( P_2 = 1 \text{ kg} \times 21 \text{ m/s} = 21 \text{ Ns} \). Since they are perpendicular, their resultant momentum \( P_{12} = sqrt{21^2 + 21^2} = 21sqrt{2} \text{ Ns} \). By conservation of momentum, \( P_3 = P_{12} = 21sqrt{2} \text{ Ns} \). Velocity of heaviest part \( v_3 = P_3 / m_3 = (21sqrt{2}) / 3 = 7sqrt{2} \text{ m/s} \).
Leave a Reply