1
K= 2mv2
Ug=mgh
Ki+Ugi = Kf +Ugf
mvi +F×t =mvf
m1v1i +m2v2i = m1v1f +m2v2f
Ballistic Pendulum
A ballistic pendulum is sometimes used in laboratories
to measure the speed of a projectile, such as a bullet.
The ballistic pendulum shown consists of a block of
wood(m2=2.5kg) suspended by a wire of negligible
mass. A bullet of mass m1=0.01kg is fired with a
speed v1i. Just after the bullet collides with it, the
block (with the bullet in it) has a speed of vf and then
swings to a maximum height of 0.65m above the
initial position. Find the speed v1i of the bullet,
assuming that air resistance is negligible.
a) 896m/s
b) 754m/s
c) 545m/s
d) 435m/s
(a)
m1
m2
v1i
(b)
h=0.650m
m1+m2
vf
Sol) We begin by writing
m1v1i +m2v2i = m1v1f +m2v2f
Since the block was stationary before the collision( v2i=0) and the bullet became embedded in the
block, so we can modify the above equation as
m1v1i = (m1 +m2 )vf
Our goal is to find v1i, but we need to find vf first in order to do that.
Since vf is the speed of the bullet+block right after the collision and the bullet+block
moves as a pendulum motion, we can use conservation of mechanical energy to find vf.
Ki+Ugi = Kf +Ugf
If we choose the position of the bullet+block the moment after the collision as the lowest point
and since when the bullet+block momentarily stops at h=0.65m, we can set Ugi = Kf =0.
Ki = Ugf
½(m1+m2)vf 2 = (m1+m2)gh
=> vf = √2𝑔ℎ = √(2 × 9.8 × 0.65) = 3.57m/s
Coming back to m1v1i = (m1 +m2 )vf, we can now solve for v1i.
0.01× v1i = (0.01 + 2.5)×3.57
v1i = 896m/s