Question

A bag of sand of mass m is suspended by a rope. A bullet of mass $\frac{m}{20}$ is fired at it with a velocity v and gets embeddedinto it. The velocity of the bag finally is(a) $\frac{V}{20}\times 21$(b) $\frac{20V}{21}$(c) $\frac{V}{20}$(d) $\frac{V}{21}$

Answer 1

According to momentum conservation equation

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Here m₁ = bullet mass

m₂ = mass of sand bag

assuming no sand is spilled during collision and bullet was fully embedded till it reached center of mass

so

u₁ = v

u₂ = 0 (as bag was initially at rest)

and v₁=v₂(both are moving with same velocity after collision)

m₁v=(m₁ + m₂)v'

so

v' = m₁v / (m₁ + m₂)

v' = m/20 * v / (m + m/20)

v' = mv/20 / (21m/20)

v' = mv/20 * 20/21m

v' = v/21 (d)