Quiz
Q1) A ballistic pendulum is a device used to measure the speed of a bullet. The wooden pendulum bob in such a system absorbs a bullet of known mass moving in a horizontal trajectory. The bullet’s speed can be calculated using the measured angle through which the pendulum swings. Most of the mechanical energy of the flying bullet captured by the pendulum bob in such a device is converted, almost instantly, into
A. mechanical energy of the swinging pendulum bob
B. internal energy of the pendulum bob and bullet
C. sound
Answer) B.
A. mechanical energy of the swinging pendulum bob
B. internal energy of the pendulum bob and bullet
C. sound
Answer) B.
Q2) A wood block rests at rest on a table. A bullet shot into
the block stops inside, and the bullet plus block start sliding on the
frictionless surface. The horizontal momentum of the bullet plus block
remains constant
A. Before the collision B. During the collision C. After the collision
D. All of the above E. Only A and C above
Answer) D.
As long as there are no external forces acting on the system
D. All of the above E. Only A and C above
Answer) D.
As long as there are no external forces acting on the system
Ballistic Pendulum Measurement Of Speed Simulation(Virtual Experiment)
When the next simulation is not visible, please refer to the following link.
Ballistic Pendulum Measurement Of Speed
The ballistic pendulum is used to measure bullet speeds. The pendulum is a large wooden block of mass $\ M $ hanging vertically by two cords. A bullet of mass $\ m $, traveling with a horizontal speed $\ v_{i}$ strikes the pendulum and remains embedded in it.
If the collision time (the time required for the bullet to come to rest with respect to the block) is very small compared to the time of swing of the pendulum, the supporting cords remain approximately vertical during the collision.
Therefore, no external horizontal force acts on the system (bullet + pendulum) during collision, and the horizontal component of momentum is conserved. The speed of the system after collision $\ v_{f}$ can be easily determined, so that the original speed of the bullet can be calculated from momentum conservation
If the collision time (the time required for the bullet to come to rest with respect to the block) is very small compared to the time of swing of the pendulum, the supporting cords remain approximately vertical during the collision.
Therefore, no external horizontal force acts on the system (bullet + pendulum) during collision, and the horizontal component of momentum is conserved. The speed of the system after collision $\ v_{f}$ can be easily determined, so that the original speed of the bullet can be calculated from momentum conservation
Figure 1. Ballistic pendulum
The initial momentum of the system is that of the bullet $\ mv_{i}$, and the momentum of the system just after the collision is
$\ (m + M)v_{f} $
Conservation of horizontal linear momentum gives
$\ mv_{i} = (m + M)v_{f} $
After the collision is over, the pendulum and bullet swing up to a maximum height y, where the kinetic energy left after impact is converted into gravitational potential energy. Then, using the conservation of mechanical energy for this part of the motion, we obtain
$\frac{1}{2}(m+M)v_{f}^2 = (m+M)gy$
Solving these two equations for $\ v_{i}$, we obtain
$\ v_{i} = \frac{m+M}{m} \sqrt{2gy}$
Hence, we can find the initial sped of the bullet by measuring $\ m $, $\ M $, and $\ y $.
If the maximum angle the pendulum swings through is θ and the length of pendulum from the center of mass to the point of rotation is $\ L $ , then the height h is given by
$\ y = L(1- cosθ) $
Figure 2. Ballistic pendulum experiment.
It would be difficult to measure the change in height $\ y $ directly. Instead, we will measure the horizontal distance $\ x $ that the pendulum moves, and use this quantity to calculate y. If we consider Figure 2, it is clear the the string length $\ L $ is the hypotenuse of a right triangle with sides $\ (L − y) $ and $\ x $, so:
$\ L^2 = (L-y)^2 + x^2 $
If we subtract $\ x^2 $ from both sides we obtain
$\ L^2 - x^2 = (L-y)^2 $
If we now take the (positive) square root and solve for $\ y $, we find
$\ y = L - \sqrt{L^2 - x^2}$
We will use this expression for $\ y $ in $\ v_{i} = \frac{m+M}{m} \sqrt{2gy}$ above to calculate $\ v_{i} $.
We also want to find the uncertainty in the initial projectile velocity $\ v_{i} $. The measured quantities, $\ x $, $\ L $, $\ m $, and $\ M $ all have some uncertainty, leading to uncertainty in the calculated $\ v_{i} $. The formulas connecting $\ v_{i} $ with these measured quantities would make the error analysis complicated. In addition, another source of uncertainty is in the behavior of the launcher—one sometimes sees variations in $\ v_{i} $ from shot to shot.