A simple pendulum when reaches its maximum angular displacement of 60° from the vertical is hit by a
bullet moving horizontally with velocity 200 m/s. The bullet comes out of the bob horizontally with
velocity 100 m/s. Find velocity of the bob immediately after the bullet comes out. Neglect loss of mass
of the bob due to piercing by the bullet. Mass of the bob is 10 times of that of the bullet.
bullet O bob
Answers
Answer:Part A
Find an expression for v, the initial speed of the fired object.
Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.
This is a purely algebraic manipulation. Solving this requires a couple of steps. The first thing to do is figure out a relationship between the initial and final velocities. Since momentum is conserved, we can use that as a starting point:
p 0 = m v 0
pf = (m+M)v
Since momentum is conserved, we can say that:
p 0 = pf
Therefore:
m v 0 = (m+M)v
and:
v 0 = ((m+M)v / m)
So the only thing left now is figure out an expression for v f. We can do this by analyzing the relationship between kinetic and potential energy:
We know that once point after the bullet hits the ballistic pendulum, the bullet and pendulum mass will fuse and the combined mass will take on kinetic energy. As the mass moves upwards, gravity will slow it down and some of the kinetic energy will be converted to potential energy. When the mass is at it’s maximum height, potential energy will be maximum and kinetic energy will be zero (gravity will have slowed the velocity to zero and the height will be maximized). Vice-versa, when the pendulum is at the bottom (zero height), potential energy will be zero and kinetic energy will be maximum.
We also know that the formula for kinetic energy has velocity in it – but we need a way to remove velocity from our expression. Potential energy doesn’t have velocity in the formula, and we know that at some point, when the pendulum is at a certain height, kinetic energy and potential energy will be equal. So we can set the two energies equal to each other, solve for velocity (in the kinetic energy side), and plug the result into our expression above:
U = (m+M)g h
K E = 1/2(m+M)v 2
(m+M)g h = 1/2 m v 2
v = s q r t(2 g h)