Question

A small wooden block with mass 0.775 kg is suspended from the lower end of a light cord that is 1.68 m long. The block is initially at rest. A bullet with mass 0.0108kg is fired at the block with a horizontal velocity . The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on theend of the cord. When the block has risen a vertical height of 0.775 m, the tension in the cord is 4.74 N.

What was the initial speed (V_0) of the bullet?

Answer 1

cosθ = (1.68-0.775)/1.68 = 57.4 degrees

T -(m+M)gcosθ =(m+M)v²/r

4.74-(07858)*9.8*cos57.4 = (0.7858)v²/1.68

v=1.124 m/s

mu²/2 =(M+m)gh + (m+M)v²/2

solve for u we get u =34.6 m/s