A ballistic pendulum is used to measure the speed of bullets. It comprises a hea

Question

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 horizontally. 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 44.0 g is fired into a wooden block of mass 11.6 kg. The block–bullet combination is observed to rise to a maximum height of 20.4 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.)

=

Explanation / Answer

Given

mass of bullet m = 44 g = 0.044 kg ,

mass of block M = 11.6 kg

height reached h = 20.4 cm = 0.204 m

by conservation of momentum

mv1 +M*v2 = (m+M)V

0.044*v1 = (11.6+0.044)V

   v1 = (11.6+0.044)V /(0.044) ------(1)

by conservation of energy

   (m+M)gh = 0.5*(m+M)V^2

   V= sqrt(gh)

   V = sqrt(9.8*0.204) m/s

   V = 1.413931 m/s
substitute V in equation (1)

   v1 = (11.6+0.044)(1.413931) /(0.044) m/s

   v1 = 374.178 m/s

a) initial speed of the bullet is v1 = 374.178 m/s

b) the kinetic energy of the bullet k.em = 0.5*m*v1^2 = 0.5*0.044*374.178^2 J = 3080.2 J

kinetic energy of the block and bullet is k.e(M+m) = 0.5*(11.6+0.044)* 1.413931^2= 11.64 J

the fraction of k.e lost after the bullet is embeded in the block is

   = k.e m - k.e(M+m) / (k.em)

   = (3080.2 -11.64)/(3080.2)

   = 0.996221

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