Question

In forensic science, it is useful to measure the speed at which firearms propel bullets. Before modern instruments, this was done with a device called a “ballistic pendulum”, consisting of a block of some soft material hanging from the end of a string. The experimenter fires a bullet into the block, which lodges into it; the block swings up at an angle. By measuring the angle, the experimenter can determine the velocity of the bullet.

Suppose that you are a detective trying to measure the velocity of the bullets fired from a particular gun. You construct a ballistic pendulum out of a string of length 50 cm and a clay block of mass 2 kg, and fire a bullet into it. If the bullet has a mass of 2.6 grams and the pendulum swings up to an angle 13.4 degrees, how fast was the bullet traveling when it struck the block?

Show all work and Thanks in advance!

Answer 1

The collision of the bullet and block will be completely inelastic because they "stick" together after the collision

Before the collision the block simply hangs without moving, hence its initial velocity will be 0, u_b = 0 m/s

Let the initial velocity of the bullet and the final velocity of the system(bullet+block) be u_g and v respectively, then

conservation of momentum gives,

m_gu_g + m_bu_b = (m_g + m_b)*v---------(1)

Let (m_g + m_b ) = M

Now, after the collision, the kinetic energy of the system is converted to its potential energy as it rises upwards, let it rise to a height 'h',

Mgh = 0.5Mv²

v =sqrt(2gh)-------(2)

then from geometry, (L*cos()+h) = L

h = L(1- cos())

putting this value in equation 2

v =sqrt(2gL(1- cos())) = sqrt(2*9.8*0.5(1- cos(13.4))) {L = 50 cm = 0.5 m}

v= 0.516 m/s

putting this in equation 1

m_gu_g + m_bu_b = (m_g + m_b)*v

2.6*10^-3*u_g + 2*0 = (2.6*10^-3 + 2)*0.516

u_g = 397.44 m/s

This is the speed of the bullet.