A ballistic pendulum uses a block of wood and a spring to calculate the initial

Question

A ballistic pendulum uses a block of wood and a spring to calculate the initial speed of a bullet (see figure). A 35.0 g bullet is fired into a 2.10 kg block that is initially at rest. The block/bullet combo swings up and strikes the spring (k = 180.0 N/m). Find the angle, theta, the pendulum swings through if the length of the pendulum is 3.50 m, the spring is 2.50 m before it is compressed, and the gun has a maximum horizontal range of 50.0 km. Assume the block is a point mass at the end of the pendulum so you can ignore the block thickness.

Explanation / Answer

maximum horizontal range = u^2 / g = 50 x 10^3 m

u = 700.36 m/s   ........speed of bullet

using momentum conservation on bullet-block mass system to find speed of block after impact,

0.035 x 700.36 + 2.10 x 0 = ( 2.10 + 0.035)v

v = 11.48 m/s

suppose spring stretches by y distance then using energy conservation,

(2.10 +0.035) x 11.48^2 /2 = mg(h+y) + ky^2 /2

140.69 = 2.135 x 9.81 x (1 + y) + 180y^2 /2

90y^2 + 20.94y - 119.75 = 0

y = 1.04 m

so block goes to (1 + 1.04) = 2.04 m high .

h = L - Lcos@

2.04 = 3.50 - 3.50cos@

3.50cos@ = 1.46

@ =cos^-1 (1.46 / 3.50)   = 65.35 deg

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