EXAMPLE 6.5 The Ballistic Pendulum GOAL Combine the concepts of conservation of energy and conservation of momentum in inelastic collisions. 2 g = g my + My & = = f s " Vii Vs = 1 it o h ; _____ & B o g (2) Diagram of a ballistic pendulum. Note that V.v_ is the velocity of the system just after the perfactly inelastic collision. (b} Multiflash photograph of a laboratory ballistic pendulum. :, 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 4. It is possible to obtain the initial speed of the bullet by measuring & and the two masses. As an example of the technigue, assume that the mass of the bullet, iy, is 5.00 g, the mass of the pendulum, M is 1.000 kg, and & is 5.00 cm. {a) Find the velocity of the system after the bullet embeds in the block. {(b) Calculate the initial speed of the bullet. STRATEGY Use conservation of energy to find the initial velocity of the block-bullet system, labeling it Ve Part (b) requires the conservation of momentum equation, which can be solved for the initial velocity of the bullet, Vi SOLUTION (A) Find the velocity of the system after the bullet embeds in the block. Apply conservation of energy to the (KE + PE) sfter collision = (KE + PE]tDp block-bullet system after the collision. Substitute expressions for the kinetic 1 2 _(m, + m,)v +0=0+(m, + m,)gh and potential energies. Mote that both 20 1 20" sys 1 2 the potential energy at the bottom and the kinetic energy at the top are zero. Solve for the final velocity of the block- Vsz = 2gh bullet system, Veyst e T . . Veys = V200 = v/ 2(9.80 m/s%)(5.00 x 1072 m) Veys = 0.2990 m/s (B) Calculate the initial speed of the bullet. Write the conservation of momentum B; = pg equation and substitute expressions. My + MoV, = [ml + mzjus Solve for the initial velocity of the {ml - mz)vs . Vy,, = > & SyS bullet, and substitute values. 1i m, 1.005 kg)(0.990 m/s vy = % = 199 m/s 5.00x 10 " kg LEARN MORE REMARKS Because the impact is inelastic, it would be incorrect to equate the initial kinetic energy of the incoming bullet to the final gravitational potential energy associated with the bullet-block combination. The energy isn't conserved! QUESTION The ways that mechanical energy is lost from the system in this experiment include: (Select all that apply.) [ friction in the mechanisms [l friction to bring the bullet to a stop relative to the ground [J emission of sound waves [ energy loss from the change in height of the block [} thermal energy loss due to air drag 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 &. It is possible to obtain the initial speed of the bullet by measuring & and the two masses. As an example of the technique, assume that the mass of the bullet, "y, is 5.48 g, the mass of the pendulum, Mo, is 1,087 kg, and % is 5.08 cm. (a) Find the velocity of the system after the bullet embeds in the block. myfs (b) Calculate the initial speed of the bullet. mys EXERCISE HINTS: GETTING STARTED | [I'M STUCK! A bullet with mass 4.86 g is fired horizontally into a 2.045-kg block attached to a horizontal spring. The spring has a constant 5.75 X 102 N/m and reaches a maximum compression of 6.27 cm. (a) Find the initial speed of the bullet-block system. myfs (b) Find the speed of the bullet. myfs Exercise . 2 al Yes, I overlooked Something. I didn't tone i. Insteed i took R. Kinetic energy = springs potential every the ( my t m2 ) U 2 = b/ikn / my = 4. 86 g = 0 .00 4 86 kg (0. 0 0 4 86 + 2 095 ) 6 2 - mz = 2 . 045 kg 5 75 X 10 * (0.0627) K = 5. 75 X 10 N/M or, 2 . 09 9 8 % ( 4 0) - 2 . 2 6 0 4 9 DK = 6127( m or 2 - 2 6 0 49 20 0627 m. 2 2 1 09 9 81 1. 10275 1:05 m/s -. initial speed of the bunet- block system - 1:05 myJ. ( 5 ) it speed the bunet be u them, mit me ) US 10.00 986 + 2095 ) x 1 05 0. 00 98% -: u = 4 42.87 my/s : inined speed of bunet 29 92. 87 MS