Question

Problem 6 (20 points) A ballistic pendulum is a device often used for measuring the velocity of a projectile. Assume to projectile fully embeds itself in the pendulum. If the projectile mass is 25 grams, the pendulum mass is 2.5 kg, and the vertical displacement of the pendulum after the collision is 12 cm, find the velocity of the projectile just prior to the collision.

Answer 1

Answer:

The velocity of the projectile just prior to the collision is 154.934 meters per second.

Explanation:

Let consider the projectile-pendulum system as our system of study, from statement we know that projectile gives kinetic energy to the pendulum and turns into gravitational potential energy. By Principle of Energy Conservation we construct the following expression:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}$ (Eq. 1)

Where:

$U_{g,1}$, $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

$K_{1}$, $K_{2}$ - Initial and final kinetic energies, measured in joules.

By applying definitions of gravitational potential and kinetic energies, we expand and simplify the equation above:

$(m+M)\cdot g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (m+M)\cdot (v_{1}^{2}-v_{2}^{2})$

$g\cdot (y_{2}-y_{1})=\frac{1}{2}\cdot (v_{1}^{2}-v_{2}^{2})$

And we clear the initial velocity of the pendulum-projectile system:

$v_{1}=\sqrt{v_{2}^{2}+2\cdot g\cdot (y_{2}-y_{1})}$ (Eq. 2)

Where:

$m$ - Mass of the projectile, measured in kilograms.

$M$ - Mass of the ballistic pendulum, measured in kilograms.

$v_{1}$, $v_{2}$ - Initial and final speeds of the pendulum-projectile system, measured in meters per second.

$g$ - Gravitational acceleration, measured in meters per square second.

$y_{1}$, $y_{2}$ - Initial and vertical position of the pendulum-projectile system, measured in meters.

Under the consideration of inellastic collision, we find the velocity of the projectile prior to the collision by Principle of Linear Momentum Conservation:

$m\cdot v_{o, P} + M\cdot v_{o,B} = (m+M)\cdot v_{1}$

$v_{o,P} = \frac{(m+M)\cdot v_{1}-M\cdot v_{o,B}}{m}$ (Eq. 2)

Where:

$v_{o, P}$ - Velocity of the project prior to the collision, measured in meters per second.

$v_{o,B}$ - Velocity of the ballistic pendulum prior to the collision, measured in meters per second.

If we know that $v_{2} = 0\,\frac{m}{s}$, $g = 9.807\,\frac{m}{s^{2}}$, $y_{2}-y_{1} = 0.12\,m$, $m = 0.025\,kg$, $M = 2.5\,kg$ and $v_{o,B} = 0\,\frac{m}{s}$, then the velocity just prior to the collision is:

By (Eq. 1):

$v_{1} =\sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.12\,m)}$

$v_{1} \approx 1.534\,\frac{m}{s}$

By (Eq. 2):

$v_{o,P} = \frac{(0.025\,kg+2.5\,kg)\cdot \left(1.534\,\frac{m}{s} \right)-(2.5\,kg)\cdot \left(0\,\frac{m}{s} \right)}{0.025\,kg}$

$v_{o,P} = 154.934\,\frac{m}{s}$

The velocity of the projectile just prior to the collision is 154.934 meters per second.