While swinging up H = 0.45 m elevation, the block and bullet convert kinetic energy to potential energy, without heat loss. This tells you that
V = sqrt(2gH) = 2.97 m/s
is the velocity of the block and bullet just after impact.
Next, apply conservation of momentum to the inelastic bullet-impact process, assuming that the pendulum does not swing during that brief interval. Let v be the bullet speed, m the bullet mass and M be the block's mass. Momentum conservation tells you that
m v = (M + m) V
Solve for v
A ballistic pendulum consists of a rectangular wooden block of mass 1.5 kg suspended from a ceiling. A bullet mass 0.025 kg undergoes a completely inelastic collision with the ballistic pendulum. Due to the collision the pendulum rises to a height of 45 cm. Calculate the speed with which bullet hits the pendulum.
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CHICKEN NUGGETS = GOLDEN TAR
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