Question

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 595 m/s at a block of wood and passes completely through it. The speed of the block is 15 m/s immediately after the bullet exits the block. (a) Determine the speed (in m/s) 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 less than the initial kinetic energy greater than the initial kinetic energy (c) Verify your answer to part (b) by calculating the initial and final kinetic energies of the system in joules. KE;= KEF = Additional Materials Reading

Answer 1

Given is:-

mass of the bullet  

m = 4.67 x 10-3kg

mass of the soft wood

M = 0.070kg

Initial speed of the bullet   

u= 595m/s

Final speed of the soft wood block  

V = 15m/s

Now,

Part - a

applying the law of conservation of momentum, we get

mu = MV + mu

by plugging all the values we get

(4.67 x 10-3kg)(595m/s) = (0.070kg)(15m/s) + (4.67 x 10-3kg)

which gives us the final velocity of the bullet, which is

v = 370.161m/s

this is the speed of the bullet when it exits the wooden block

part - b,c

Initial kinetic energy

KE; = 5mu

or

KE; = }(4.67 x 10-3kg)(595m/s) ?

KE; = 826.648J

Similarly

KE; = 5mo? + mva

or

KE; = (4.67 x 10-3kg)(370.161m/s)2 + (0.070kg)(15m/s)?

which gives us

KE, = 327.815J

We can clearly see that the final kinetic energy is less than the initial kinetic energy