Explosion and Momentum Conservation – Rankers Physics
Topic: Center of Mass , Momentum and Collision
Subtopic: Collision

Explosion and Momentum Conservation

An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, \( 1 \text{ kg} \) first part moving with a velocity of \( 12 \text{ m s}^{-1} \) and \( 2 \text{ kg} \) second part moving with a velocity of \( 8 \text{ m s}^{-1} \). If the third part flies off with a velocity of \( 4 \text{ m s}^{-1} \), its mass would be:

(2009)

\( 7 \text{ kg} \)
\( 17 \text{ kg} \)
\( 3 \text{ kg} \)
\( 5 \text{ kg} \)

Solution:

By conservation of momentum, \( P_{text{total}} = 0 \). Momentum of first part \( P_1 = 1 \text{ kg} \times 12 \text{ m/s} = 12 \text{ Ns} \). Momentum of second part \( P_2 = 2 \text{ kg} \times 8 \text{ m/s} = 16 \text{ Ns} \). As \( P_1 \) and \( P_2 \) are perpendicular, their resultant \( P_{12} = sqrt{12^2 + 16^2} = 20 \text{ Ns} \). For conservation, \( P_3 \) must be \( 20 \text{ Ns} \). \( m_3 = P_3 / v_3 = 20 \text{ Ns} / 4 \text{ m/s} = 5 \text{ kg} \).

Leave a Reply

Your email address will not be published. Required fields are marked *