Question

In a ballistic pendulum, a spring pushes a ball from rest It ies through the air and sticks in the base of a pendlum that swings upwards. Which of the following equations correctly represents the velocitye of the pendulum after the ball has collided with i? A. E. 5. In a balitic pendulum, a spring pushos a ball from rest t lles through the air and sticks in the base of a pendlum that swings upwards Given that the ball and pendulum reach a maximum angle of 45 degrees, the pendulum is 30 cm long, the mass of the ball is 76 g, and th mass of the pendulum is 250 g, calculate the speed of the ball after it has left the spring but before it hits the pendulum. Do not include th units in your response.

Answer 1

Answer:

The final velocity is $\bf{562.9}$.

Explanation:

Given:

The maximum angle that the pendulum and the ball system can travel, $\theta_{m} = 45^{0}$.

The length of the pendulum, $l = 30~cm$.

The mass of the pendulum, $m_{p} = 250~g$.

The mass of the ball, $m_{b} = 76~g$.

Consider that the initial velocity of the ball-pendulum system is $v_{i}$. So the initial total energy of the system is given by

$E_{i} = K.E. + P.E.\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} + 0\\~~~~= \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2}$

Consider the final velocity of the ball is $v_{0}$ and the final height attended by the system is $h$. So the final total final energy of the system is given by

$E_{f} = K.E. + P.E.\\~~~~~= 0 + (m_{b} + m_{s})gh\\~~~~~= (m_{b} + m_{s})gh$

From the conservation of energy,

$E_{i} = E_{f}\\&or,& \dfrac{1}{2}(m_{b} + m_{s})v_{i}^{2} = (m_{b} + m_{s})gh\\&or,& v_{i} = \sqrt{2gh}$

From the conservation of momentum,

$&& m_{b}v_{0} = (m_{b} + m_{s})v_{i}\\&or,& v_{0} = (1 + \dfrac{m_{p}}{m_{b}})\sqrt{2gh}$

The final height attended by the system is given by

$h = l(1 - \cos 45^{0}) = 8.787~cm$

The final velocity is given by

$v_{0} = (1 + \dfrac{250}{76}})\sqrt{2(980)(8.787)}\\~~~~= 562.9$