Exploding Mass in Projectile Motion – Rankers Physics

Collision: Practice Problem & Solution

A mass of \( 1 \text{ kg} \) is thrown up with a velocity of \( 100 \text{ m/s} \). After \( 5 \) seconds, it explodes into two parts. One part of mass \( 400 \text{ g} \) comes down with a velocity \( 25 \text{ m/s} \). Calculate the velocity of other part: (2000)
\( 40 \text{ m/s} \) upward
\( 40 \text{ m/s} \) downward
\( 100 \text{ m/s} \) upward
\( 60 \text{ m/s} \) downward

Solution Explained:

To solve this problem, we apply the core principles of Collision. Understanding the underlying formula is key to arriving at the correct answer below:

Velocity of \( 1 \text{ kg} \) mass after \( 5 \text{ s} \): \( v = u - gt = 100 - 10 \times 5 = 50 \text{ m/s} \) (upward). Initial momentum before explosion \( P_i = 1 \text{ kg} \times 50 \text{ m/s} = 50 \text{ Ns} \) (upward). Mass of first part \( m_1 = 0.4 \text{ kg} \), \( v_1 = -25 \text{ m/s} \). Mass of second part \( m_2 = 0.6 \text{ kg} \). By conservation of momentum: \( P_i = m_1 v_1 + m_2 v_2 \). \( 50 = 0.4 \times (-25) + 0.6 v_2 \). \( 50 = -10 + 0.6 v_2 \implies v_2 = 100 \text{ m/s} \) (upward).

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