A bullet (m = 10 g) flies horizontally with an initial speed of v = 500...

A bullet (m = 10 g) flies horizontally with an initial speed of v = 500 m/s and then embeds itself in a block of wood (m = 400 g) that is at rest on a flat surface. The surface has a coefficient of kinetic friction μk = 0.4. After the impact, how far does the block of wood slide?

(a) 40 m (b) 800 m (c) 8 m (d) 80 m (e) 20 m

Expert Solution

According to the momentum conservation, the momentum of the bullet moving with a velocity v before the embedding is equal to the momentum of the combined (bullet + wood block) system which is assumed to be moving at the velocity u immediately after the impact. And so, symbolically, we get

So, the kinetic energy of the combined system is

And according to the friction law, the work done against the friction force in moving a distance x is given by

And the combined system will cease to slide when all of its kinetic energy is lost due to work done against the force of friction. So, we get



Now, putting the given values,

we get the distance travelled


This answer is very close to the option (e) 20 m.
So, option (e) is correct.

genius_generous answered 8 months ago

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