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Vesnalui [34]
3 years ago
13

A 0.0220 kg bullet moving horizontally at 400 m/s embeds itself into an initially stationary 0.500 kg block. (a) What is their v

elocity just after the collision?
(b) The bullet-embedded block slides 8.0 m on a horizontal surface with a 0.30 kinetic coefficient of friction. Now what is its velocity?
(c) The bullet-embedded block now strikes and sticks to a stationary 2.00 kg block. How far does this combination travel before stopping?
Physics
1 answer:
nevsk [136]3 years ago
5 0

Answer:

a) 16.86 m/s.

b) 15.40 m/s

c) 3.175 m/s  

Explanation:

let pi be the initial momentum of the system, pf be the final momentum of the system, m1 be the mass of the bullet and V1 be the intial velocity of the bullet, m2 be the mass of the block and V2 be the intial velocity of the block

a)  from the conservation of linear momentum:

                                          pi = pf

              m1×(V1)i + m2×(V2)i = Vf×(m1 + m2)

(0.0220)×(400) + (0.500)×(0) = Vf(0.0220 + 0.500)

                                         8.8 = 0.522×Vf

                                          Vf = 16.86 m/s

Therefore, the bullet-embedded block system will be moving with a speed of 16.86 m/s.

b)  let Vf be the final velocity of the bullet-embedded block system, Wf be the work done by friction on the system, Vi be the initial velocity that the bullet-embedded block system initially moves with.

the total work done on the system is given by:

                                Wtot = Δk = kf - ki

                                    Wf = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                    f×ΔxΔcos(Ф) = 1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                 m×Δx×g×μ×(-1) =  1/2×m×(Vi)^2 - 1/2×m×(Vf)^2

                        m×g×Δx×μ =  1/2(Vi)^2 - 1/2(Vf)^2

1/2(0.0220 + 0.500)(Vf)^2 = 1/2(0.0220 + 0.500)(16.86)^2 - (0.0220 +           0.500)×(9.8)×(8)×(0.30)

                     (0.261)(Vf)^2 = 86.469

                                (Vf)^2 = 237.2196

                                       Vf = 15.40 m/s

Therefore,   bullet-embedded block system will have a speed of 15.40 m/s after sliding of the friction surface.

c) let Vf be the speed of the bullet-embedded before hitting the block, Vi be the initial velocity of the block and V be the velocity of the bullet-embedded block and block system.

                              M×Vf + m×Vi = (m + M)×V  

 (0.0200 + 0.50)×(15.40) + (2)×(0) = (0.022 + 0.50 + 2)×V

                                             8.008 = 2.522×V

                                                    V = 3.175 m/s  

  Therefore, the bullet-embedded block and block system will be moving at a speed of 3.175 m/s.

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Explanation:

family They don't want them . There was a lady from there neighborhood can't bearer she said I can Take care of them . She was multi Millionaire she changed their names and she brought for them bed rooms ، bathroom and thire on toy's And a special clothing . they lived with her only 5 years then she said for the oldest daughter

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After 2 weeks the step mother gaved the maid and told her.

Step mother: this is 3 millions for the kids and take care of them and don't tell them that I am traveling .

Maid : and if they asked me

step mother : she went to her family 5 months only .

Maid: but where you are going madame

Step mother: actually I am going to do a surgery in london

Maid : but why

Step mother: because I can't beCome

Pregnant

Maid: but you have mikinzy kareem and natalie

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After 2 weeks mikinzy asked the maid where is my mother the maid told her everything mikinzy started to cry the maid went to her room

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