Question

In a ballistic pendulum demonstration gone bad, a 0.53 g pellet, fired horizontally with kinetic energy 3.55 J , passes straight through the 400 g Styrofoam pendulum block. If the pendulum rises a maximum height of 0.55 mm , how much kinetic energy did the pellet have after emerging from the Styrofoam? Express your answer using two significant figures.

Answer 1

Using momentum conservation:

Pi = Pf

m1*v1i + m2*v2i = m1*v1f + m2*v2f

here, m1 = mass og pellet = 0.53 g

m2 = mass of block = 400 g

v1i = sqrt(2*KEi/m1)

v2i = 0

v1f = final speed of pellete = ??

v2f = final speed of block

Using energy conservation on block,

Kinetic energy of block = Final potential energy of block

0.5m2*v2f^2 = m2*g*h

v2f = sqrt(2*g*h)

Given, KEi = 3.55 J

h = 0.55 mm = 0.55*10^-3

So, m1*sqrt(2*KEi/m1) + m2*0 = m1*v1f + m2*sqrt(2*g*h)

Using known values:

0.53*sqrt(2*3.55/(0.53*10^-3)) + 0 = 0.53*v1f + 400*sqrt(2*9.81*0.55*10^-3)

v1f = [0.53*sqrt(2*3.55/(0.53*10^-3)) - 400*sqrt(2*9.81*0.55*10^-3)]/0.53

v1f = 37.34 m/s

So, Final kinetic energy of pellet will be:

KEf = 0.5*m1*v1f^2 = 0.5*(0.53*10^-3)*37.34^2 = 0.369

In two significant figure:

KEf = 0.37 J