Question

A 5.5 g dart is fired into a block of wood with a mass of 22.6 g. The wood block is initially at rest on a 1.5 m tall post. After the collision, the wood block and dart land 2.5m from the base of the post. Find the initial speed of the dart.

Answer 1

The law of conservation of momentum gives

mv_0=(m+M)u

Here u is the velocity of the wood block and dart after collision.

Since

l=u√(2h/g)

We get

u=l/√(2h/g)=2.5/(√(2×1.5)/9.8)=4.5 m/s

Finally

v_0=(m+M)/m u=(5.5+22.6)/5.5×4.5=23 m/s

The law of conservation of momentum gives

mv_0=(m+M)u

Here u is the velocity of the wood block and dart after collision.

Since

l=u√(2h/g)

We get

u=l/√(2h/g)=2.5/(√(2×1.5)/9.8)=4.5 m/s

Finally

v_0=(m+M)/m u=(5.5+22.6)/5.5×4.5=23 m/s