A bullet with a mass of 40 g and a speed of 640 ms is fired...

A bullet with a mass of 4.0 g and a speed of 640 m/s is fired at a block of wood with a mass of 9.7×10^(-2) kg. The block rests on a frictionless surface and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 22 m/s. Part A What is the speed of the bullet when it exits the block? Express your answer using two significant figures. v = ______ m/s Part B Is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Part C Explain. Part D Verify your answer to part B by calculating the initial and final kinetic energies of the system. Express your answer using two significant figures. Part E Express your answer using two significant figures.

Transcript

00:01 The bullet with a mass of 4 grams and a speed of 640 meters per second is fired into a block of wood with the mass of 9 .7 times 10 to the power minus 2 kilograms.

00:10 The block rests on a frictionless surface and is thin enough that the blast passes completed through it.

00:16 Nearly after the bullet exits the rock, the speed of the block is 22 meters per second, i won't find the speed of the blot when it exits the block.

00:26 So momentum is conserved.

00:30 M1, u1 plus m2.

00:33 U2, we call to m1, v1 plus m2, v2, v2.

00:40 So here one will be for the bullet and two will be for the block.

00:49 M is mass, u is the initial velocity, and v is the final velocity.

00:54 So start with the mass of the bullet we have 4 grams, 4 times 10 to power of minus 3 kilograms, convertent to kilograms times u1 which is the incher velocity of the bullet 640 plus m2 for the amount of the block 9 .7 times 10 to power minus 2 kilograms times the inch of velocity of the wood which is going to be zero since was stationary resting and m1 4 times connected the power minus 3 final velocity of the bullet or tool that's 22 meters per second as they exit exits the block plus m2 which is 9 .7 and 10 to the above minus 2 final velocity of the sorry this is unknown that's the velocity of the bullet and we know the velocity of the block has 22 meters per second so all the velocities are in the same direction so they have the same positive sign so solving this equation v1 this gives us 106 .5 per second for the velocity of blitz right up that the exits a block so that's for part a and for b okay we want to express this statistic in configures so that approximately be equal to 1 .1 which is up 2 meters per second part b is the final kinetic energy of the system equal to less than or greater than the initial kind of can inject the system so be in the shell k e we have one over two m1 u1 squared that's one over two m2 u2 squared so it's going to be 1 over 2 times 4 times 10 to the power minus 3 times 640 squared and this is going to be zero because our initial velocity is here this is a big quarter.

03:31 It gives us 819 .2 jewels.

03:42 Then for the final, k .e.

03:48 1 over 2.

03:50 This time we're going to use the final velocities.

03:52 1 .2 times 4 times 10 to the power of minus 3 times v1, which was 10 .5 square, plus 1 over 2 times 9 .7 times 10 to the power of minus 3...

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