2gh (velocity of bullet just before impact) Constants By measuring m, mw. and h,

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2gh (velocity of bullet just before impact) Constants By measuring m, mw. and h, we can compute the original velocity v of the bullet. For example, if mB h 3.00 cm 0.0300 m, then 5.00 g 0.00500 kg, mw 2.00 kg. and Now we will Icok at a classic physics lab apparatus for measuring the speed of a projectle. (Figure 1) and (Figure 2) show a simple form of the apparatus, known as a ballistic pendulum, composed of a block and some string. A bullet with mass B is fired into a block of wood with mass mw suspended like a pendulum. The bullet makes a completely inelastic collision with the block, becoming embedded in it. After the impact of the bullet, the block swings up to a maximum height h. Given values of h, MB, and m, how can we find the initial speed u of the bullet? What becomes of its initial kinetic energy? 2.00 ks+0,00502(9.80 m/s2) (0.0300 m) 0.00500kg 307 m's The z component V of velocity of the block just after impact is V V/2gh = V/2(9.80 m/s*) (0.0300 m) 0.767 m/s = Once we have the needed velocities, we can compute the kinetic energies just before and just after impact. The total kinetic energy just before impact is K(0.0050 kg) (307 m/s) 236 J. We find that just after impact, it is K)V(2.005 kg)(0.767 m/s)0.590 J. Only a small fraction of the initial kinetic energy remains REFLECT When an object collides inelastically with a stationary object that has a much larger mass, nearly all of the first object's kinetic energy is lost. In this problem, the wood splinters, and the bullet and wood become hotter as mechanical energy is converted to internal energy. Part A Practice Problem Suppose the mass of the bullet, the mass of the block, and the height of the block's swing have the same values as above. If the bullet goes all the way through the block and emerges with one-sixth its intial velocity, what was its initial speed? Express your answer to three significant figures and include appropriate units Figure of2 > ValueUnits Before Collision Submit quest Answ PB myy Return to Assignment Provide Feedback

Explanation / Answer

Apply conservation of momentum

mBv= mw vw + mB * v/6

0.005 v = 2 vw + 0.005 ( v/6)

vw = v/480

Apply conservation of energy between the lowest and highest point of the pendulum

PEi + KEi = PEf + KEf

0+ 1/2 * (2) ( v/ 480)^2 = 2 (9.8) ( 0.03)+0

v= 368 m/s

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