Notes O Ask Your Teacher 7. 1 points OSColPhys2016 8.5.WA.038 A thin block of so

Question

Notes O Ask Your Teacher 7. 1 points OSColPhys2016 8.5.WA.038 A thin block of soft wood with a mass of 0.070 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is fired with a speed of 629 m/s at a block of wood and passes completely through it. The speed of the block is 20 m/s immediately after the bullet exits the block. (a) Determine the speed of the bullet as it exits the block. m/s (b) Determine if the final kinetic energy of this system (block and bullet) is equal to, less than, or greater than the initial kinetic energy O equal to the initial kinetic energy O less than the initial kinetic energy O greater than the initial kinetic energy (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system. KEI KEf = Additional Materials Reading

Explanation / Answer

a)

This collision is inelastic. In this type of collision, momentum is conserved but kinetic energy is not. From the law of conservation of momentum (linear) :

mv(i) +mv(i) = mv(f) + mv(f)

Since the block is initially at rest, mv(i) = 0. Therefore :
mv(i) = mv(f) + mv(f)

Solving for v(f) :

v(f) = [mv(i) - mv(f)] / m
= [(0.00467 kg)(629 m/s) – (0.070 kg)(20 m/s)] / 0.00467kg
= 329.21 m/s

c)

The kinetic energy of the system is nothing but KE of the bullet before the collision is :

KEi = 0.5mv²
= 0.5(0.00467 kg)(629 m/s)²
= 923.82 J

after the collision the bullets

KEf = 0.5*(0.00467 kg)*(329.21 m/s)²  = 253.06 J

b) from option c final Kinetic energy is Less than initial Kinetic energy

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