Question

A 5.20g bullet moving at 672 m/s strikes a 700g wooden block atrest on a frictionless surface. The bullet emerges, travelingin the same direction with its speed reduced to 428 m/s.
a. What is the resulting speed of the block?
b. What is the speed of the bullet-block center of mass?

Answer 1

We have:
Mass of bullet m_b = 5.20 g
Speed of bullet initial (V_i)_b = 672 m/s
mass of wooden block m_bl = 700g
speed of block initial (V_i)_bl = 0
final speed of bullet (V_f)_b = 428m/s

final speed of block (V_f)_bl = ?
given by law of conservation of momentum
m_b * (V_i)_b + (m_bl)*(V_i)_bl = m_b*(V_f)_b + m_bl *(V_f)_bl
5.20g * 672m/s + 700g*0 = 5.20g*428m/s +700g * (V_f)_bl
(V_f)_bl = 1.81m/s

let theV_com be the speed of the bullet block center of massso,
V_com = [m_b* (V_i)_b]/[m_b + m_bl]
          = [5.2 *672]/[5.2+700]
          = 4.955 m/s

Hope this helps

Answer 2

Actually, ^that^ is wrong. Everything is correct expect for thatthe final velocity of the bullet is 428 m/s, not (672-428) m/s. Itsays "reduced TO 428" not "reduced by 428"

So the final velocity of the block is acutally about 1.81 m/s

Answer 3

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a)

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=>     

=>   

=>      (^m/_s )