(a) As the bullet remains in the block, we can treat it as an inelastic collision.
Let 1 represent bullet and 2 represent block
m1v1 + m2v2 = (m1 + m2)v
as block was at rest initially, so we have
m1v1 = (m1 + m2)v
v = m1v1 / m1 + m2
v = 0.015 * 50 / 0.015 + 2.5
v = 0.2982 m/s
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Initial energy = 1/2 * m1 * v12 = 1/2 * 0.015 * 502 = 18.75 J
final energy = 1/2 * (m1 + m2) * v2 = 0.1118 J
so,
energy dissipated = 18.75 - 0.1118
energy dissipated = 18.638 J
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