A bullet of mass 0.053 kg traveling horizontally at a speed...

A bullet of mass 0.053 kg traveling horizontally at a speed of 200 m/s embeds itself in a block of mass 1.5 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? v = ____ m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: $K_i$ = ____ J (c) Calculate the kinetic energy of the bullet plus the block after the collision: $K_f$ = ____ J d) Was this collision elastic or inelastic? \begin{itemize} \item elastic \item inelastic \item not enough information to tell \end{itemize} (e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision: $\Delta E_{thermal, bullet} + \Delta E_{thermal, block}$ = ____ J (f) What was the transfer of energy Q (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that Q represents energy transfer due to a temperature difference between a system and its surroundings.) Q = ____ J

Transcript

00:01 This is the data of this bullet.

00:06 Let m1 is the mass of the bullet, 0 .064 kilogram.

00:16 U1 is the velocity of the bullet up before the bullet hits the block.

00:23 250 meter per second.

00:26 And this is the data of the block.

00:30 M2 is the mass of the block, 2kg, and u2 is the velocity.

00:37 Of the block before bullet hits the block is zero because the bullet was addressed.

00:45 After the collagen, the bullet gets embedded into the block and they move like a combined object.

00:54 So for part a, we use the law of conservation of linear momentum and what is the system here? the system is bullet and block.

01:08 This is the name of the system.

01:11 And what is the event where the linear momentum will change? the event is the, means bullet hit the block.

01:24 When the bullet hits the block, this is the event.

01:28 So before the event and after the event, the total linear momentum of the system should be same.

01:36 So it means before the bullet, this is the event.

01:43 The event before before bullet hits block the total linear momentum of system should be equal to after bullet hits the block the total linear momentum of the system after the hit the total linear momentum of the system after the bullet hits so we know that the linear momentum is equal to mass into velocity so linear momentum of the bullet is equal to its mass into its velocity plus linear momentum of the block is its mass m2, its velocity, u2.

02:34 And after the even they move with like a combined object.

02:39 So let capital v is the velocity of the combined object.

02:44 Combined velocity.

02:45 Let capital v is there combined velocity.

02:52 So it will be m1 capital v and plus m2 capital v.

03:01 So here now u2 is zero because the block was at rest.

03:06 So it means m1 u1 is equal to m1 plus m2 into capital v, which means capital v is equal to m1 u1 upon m1 plus m2.

03:24 We have all the values we can simply put the values m1 is 0 .064 kilogram u1 is 250 meter per second and m1 is 0 .064 kilogramm 2 kilogram so we saw this it comes out to be 7 .7519 meter per second right now in part b they are asking about the kinetic energy before the heat.

04:01 So let kinetic energy before the heat, we call it kinetic energy 1.

04:05 This is equal to the kinetic energy of bullet, right? and plus kinetic energy of this block.

04:15 And we know that kinetic energy is equal to one half md square.

04:19 So kinetic energy of the bullet will be one half, m1, u1 square.

04:25 Kinetic energy of the block will be one half m2 u2 square now here the means block is stationary u2 is zero so this will be one half m1 u1 square and m1 is the mass of the bullet 0 .0 64 kilogram into u1 is the speed of the bullet 250 meter per second this comes out to be 2 000 june right now in part c, they are asking kinetic energy 2 means kinetic energy after the collision.

05:04 Canatic energy after the collision here again, the kinetic energy of the bullet after the collision is equal to one half mass of the bullet multiplied by capital v.

05:15 Capital v is the combined velocity of the bullet and the block after code.

05:20 And for the block, it will be one half m2 capital v square.

05:24 Right so this will be one half m1 plus m2 into capital v square so one half m1 is 0 .064 plus 2 multiplied by 7 .7519 capital means a combined velocity we have just calculated in the previous part so this comes out to be 62 .0149 q...

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