Asked by joy
A ballistic pendulum is used to measure the speed of bullets. It comprises a heavy block of wood of mass M suspended by two long cords. A bullet of mass m is fired into the block in the horizontal direction. The block, with the bullet embedded in it, swings upward (see figure below). The center of mass of the combination rises through a vertical distance h before coming to rest momentarily. In a particular experiment, a bullet of mass 48.0 g is fired into a wooden block of mass 11.2 kg. The block–bullet combination is observed to rise to a maximum height of 19.6 cm above the block's initial height.
(a) What is the initial speed of the bullet?
m/s
(b) What is the fraction of initial kinetic energy lost after the bullet is embedded in the block? (Enter your answer to at least three significant figures.)
(KEm-KEM+m)/KEm=
(a) What is the initial speed of the bullet?
m/s
(b) What is the fraction of initial kinetic energy lost after the bullet is embedded in the block? (Enter your answer to at least three significant figures.)
(KEm-KEM+m)/KEm=
Answers
Answered by
scott
(a) momentum is conserved ... .0480 Vib = (11.2 + .048) Vbb
... (Vbb)^2 = 2 * g * 0.0196
(b) KEb = 1/2 * 0.0480 * (Vib)^2 ... this is KEm
... KEbb = 11.248 * g * 0.0196 ... this is KEM+m
... (Vbb)^2 = 2 * g * 0.0196
(b) KEb = 1/2 * 0.0480 * (Vib)^2 ... this is KEm
... KEbb = 11.248 * g * 0.0196 ... this is KEM+m
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