Question

A small wooden block with mass 0.750 kg is suspended from the lower end of a light cord that is 1.66 m long. The block is initially at rest. A bullet with mass 0.0128 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.775 m , the tension in the cord is 4.68 N .

What was the initial speed v0 of the bullet?

Answer 1

Solution,

Height, h = l(1 - cosx)

0.775 = 1.66(1 - cosx)

Angle, x = 57.78 degree

Radius, R = l sinx = 1.66 x sin57.78 = 1.40 m

T - mg cosx = mv^2/R

4.68 - (0.75 + 0.0128) 9.8 x cos57.78 = (0.75 + 0.0128) v^2/1.4

Velocity, v = 1.13 m/s

From law of conservation of energy

u = sqrt[2 (gh + v^2/2]

u = sqrt[2 x ((9.8 x 0.775) + 1.13^2/2) = 4.06 m/s

Using law of conservation of momentum

(0.775 x 0) +(0.0128 x vo) = (0.775 + 0.0128) 4.06

Velocity, vo = 249.73 m/s

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