HMWK T Practice Problem 8.7 a previous | 7 of 19 Inext Practice Problem 8.7 SOLV

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HMWK T Practice Problem 8.7 a previous | 7 of 19 Inext Practice Problem 8.7 SOLVE (Eigure 1) shows the situation just before impact: (Figure 2) shows the situation just after impact, and the block (with the embedded bullet) at the highest point of its path. Conservation of momentum just before impact and just after gives us Now we will look at a classic physics lab apparatus for measuring the speed of a projectile. (Fiqure 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 TTR 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 mw, how can we find the initial speed u of the bullet? What becomes of its initial kinetic energy? The kinetic energy of the system just after the collision is K)V. The pendulum comes to rest (for an instant) at a height h, where its kinetic energy (mB + mw )V2 has all become potential energy (mig + mw)gh; then it swings back down Energy conservation gnes Assuming that h can be measured, we solve this equation for V V- 2gh (velocity of block and bullet just after impact). Now we substitute this result into the momentum equation to find w: u = m" my V2gh (velocity of bullet just before impact) By measuring mBmw. and h, we can compute the original velocity u of the bullet. For example, if mB = 5.00 g = 0.00500 kg. mw h 3.00 cm = 0.0300 m, then 2.00 kg. and 00050 29.80 m/0.0300 m) 0.00500kg 307 m/s The a component V of velocity of the block just after impact is V2gh=v/2(9.80 m/s*)(0.0300 m) 0.767 m/s V = Figure 1 # of 2 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, mBa,2-1 (0.00500 kg) (307 m/s)2-236 J. We find that just after impact, it is Kj- (m B + mw)V 2-1 (2.005 kg) (0.767 m/s) 0.590 J . Only a small fraction of the initial kinetic energy remains. REFLECTWhen 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 Before Collision problem, the wood splinters, and the bullet and wood become hotter as mechanical energy is converted to internal energy. Part A Practice Problem: iB 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-tenth its initial velocity, what was its initial speed? Express your answer to three significant figures and include appropriate units.

Explanation / Answer

Let the velocity of bullet be v and velocity of box be V

Conserving momentum

mv=MV+mv/10

9mv/10=MV

We got

v=(10M/9m)V

For block

Initial kinetic energy=final potential energy

0.5MV*V=Mgh

We got V=0.767 m/s

v=340.8 m/s

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