Question

A bullet weighing 0.05 kg penetrates into a blockof wood of mass 2 kg with a velocity of 150m/sand gets embedded the block of wood is theblock of wood was intially at rest calculate bthe work done

Answer 1

Answer:

Explanation:

Answer:

Δp=1.4•10^-4 kg•m/s.

Explanation:

If the block doesn’t move with bullet inside, the magnitude of acceleration (deceleration) of the bullet is

a =v²/2s=350²/2•0.15=408333 m/s².

F=ma= 0.05•408333= 20417 N.

The change in momentum of the bullet

Δp=p2-p1=0-m•v=-0.05•350=-1.4•10^-4 kg•m/s.

The impulse exerted by the wood on the bullet = Δp=1.4•10^-4 kg•m/s.