Question

QUESTION The ways that mechanical energy is lost from the system in this experiment include: (Select all that apply.) friction in the mechanisms emission of sound waves energy loss from the change in height of the block thermal energy loss due to air drag friction to bring the bullet to a stop relative to the ground PRACTICE IT Use the worked example above to help you solve this problem. The ballistic pendulum (Figure a) is a device used to measure the speed of a fast-moving projectile such as a bullet. The bullet is fired into a large block of wood suspended from some light wires. The bullet is stopped by the block, and the entire system swings up to a height h. It is possible to obtain the initial speed of the bullet by measuring h and the two masses. As an example of the technique, assume that the mass of the bullet, m1, is 4.96 g, the mass of the pendulum, m2, is 1.037 kg, and h is 4.64 cm (a) Find the velocity of the system after the bullet embeds in the block m/s (b) Calculate the initial speed of the bullet. m/s EXERCISE A bullet with mass 5.12 g is fired horizontally into a 2.105-kg block attached to a horizontal spring. The spring has a constant 6.37 X 10 N/m and reaches a maximum compression of 5.80 cm HINTS: GETTING STARTED II'M STUCK! (a) Find the initial speed of the bullet-block system m/s (b) Find the speed of the bullet. m/s Need Help? Talk to a Tutor

Answer 1

the mechanical energy is lost from the system because the work is done on the system, it lost the energy due to the friction friction to bring the bullet to a stop relative to the ground m_1 = 4.96g = 4.96 times 10^-3 kg m_2 = 1.037 kg h = 4.64 cm = 0.0464 m a) upsilon_sys = squareroot 2gh upsilon = squareroot 2 times 9.81 m/s^2 times 0.0464 m upsilon_sys = 0.954 m/s b) upsilon_ib = (m_1 + m_2/m_1) times upsilon_sys upsilon_ib = [4.96 times 10^-3 kg + 1.037 kg/4.96 times 10^-3 kg] times 0.954 m/s v_ib = 200.44 m/s m_1 = 5.12 times 10^-3 kg, m_2 = 2.105 kg k = 6.37 times 10^2 N/m , x = 5.8 times 10^-2 m a) v_i = squareroot kx^2/m_1 + m_2 = squareroot 6.37 times 10^2 N/m alpha(5.8 times 10^-2 m) ^2/5.12 times 106-3 kg + 2.105 kg v_i = 1.008 m/s b) v_b = (m_1 + m_2/m_1) times upsilon^2_1 = [5.12 times 10^-3 kg + 2.105/5.12 times 10^-3 kg] times 1.008 m/s