A50 gram bullet is fired through a 5 kg block of wood resting on a frictionless surface. the bullet has an initial speed of 500 m/s, while the block is initially at rest. after passing through the block, the bullet has a speed of 300 m/s. how fast is the wooden block moving after the bullet passes through it?

v₂ = 2 m/s

Explanation:

Given that

m₁ = 50 gm , u₁= 500 m/s   ,v₁ = 300 m/s

m₂ = 5 kg   ,u₂=0      

The final speed of the block = v₂

There is no any external force in the system  that is why the linear momentum of the system will be conserved

Initial linear momentum = Final linear momentum

m₁u₁+ m₂ u₂ =m₁v₁+m₂ v₂

0.05 x 500 + 5 x 0 = 0.05 x 300 + 5 x v₂

25 + 0 = 15 + 5 v₂

25 - 15 = 5 v₂

10 = 5 v₂

v₂ = 2 m/s

So the final speed of the wooden block will be 2 m/s.  

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

where is unknown. because the position is parabolic, there is symmetry about the midpoint at , and we know that

so that the position function is

which means the velocity is

then the total distance traveled is

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at and ends at , so its displacement is 0.

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y