Question

A ballistic pendulum consists of a large heavy mass, M on the end of a very light rod of length L. The rod is free to pivot at the top and the mass is attached at the bottom, so it hangs like a pendulum. A second mass, m, is fired horizontally at speed v straight into the large mass, and they stick together. The pendulum swings to a maximum angle ?m.

a)Show that you can measure v by measuring ?m, that is obtain an expression for v in terms of M, m, L, g and ? m.

b)Look up the mass and speed of a bullet, and figure out how mass

ive M should be if L is about 0.25 m and we want a maximum angular displacement no greater than 45?.

Answer 1

a)

Using conservation of momentum for small and large mass

m = small mass

M = large mass

momentum before collision = mv + M(0)

momentum after collision = (m + M)vf

Pi = Pf

mv = (m + M)vf

v = [(m + M) / m]*vf .....................(1)

Now using conservation of energy as both object rises after the collison,

Initial energy of system is

Ei = 1/2*(m + M)vf^2 + Ui

Ui is the initial potential energy

final energy of the system is

Ef = KE + Uf

here KE = 0

Ef = Uf

Therefore, Ei = Ef

1/2*(m + M)vf^2 + Ui = Uf

Uf - Ui = 1/2*(m + M)vf^2

U = 1/2*(m + M)vf^2 ...........(2)

U = change in potential energy = (m + M)gh

h = L - L*cos(m)

U = (m + M)g*L(1 - cosm)

putting in equation (2)

1/2*(m + M)vf^2 = (m + M)g*L(1 - cosm)

vf = sqrt[(2gL*(1 - cosm)] putting in (1)

v = [(m + M) / m] x sqrt[(2gL*(1 - cosm)]

b)

I think (b) is incomlete.