Q: A bomb of 1 kg is thrown vertically up with speed 100 m/s. After 5 seconds, it explodes into two parts. One part of mass 400 gm goes down with speed 25 m/s. What will happen to the other part just after explosion
Sol: After 5 sec, velocity of the bomb,
$\large \vec{v} = \vec{u} + \vec{gt}$
$\large \vec{v} = u \hat{j} – g t \hat{j} $
$\large \vec{v} = 100 \hat{j} – 10 \times 5 \hat{j} $
$\large \vec{v} = 50 \hat{j} m/s$
m = 1kg , m1 = 0.4 kg, m2 = 0.6 kg, v1 = 25 m/s
According to law of conservation of momentum
$\large \vec{m v} = \vec{m_1 v_1} + \vec{m_2 v_2} $
$\large 1 \times 50 \hat{j} = -0.4 \times 25 \hat{j} + 0.6 \vec{v_2}$
$\large \vec{v_2} = 100 \hat{j}$ ; vertically upwards.