phys momentum的問題-1

2009-06-15 12:40 pm

有無高手可以將整個過程做比我睇?

A small block of mass M_1=4.0kg is initially at rest on the surface of a frictionless, horizontal table. A bullet of mass M_2=50g is fired horizontally into the block with initial speed V_i = 8.0 X 〖10〗^2m/s. The bullet then emerged very quickly from the other side of the block with a reduced speed V_f= 0.75V_i.
Calculate:
a)The speed of the block after it is penetrated by the bullet.
b)The amount of mechanical energy of the bullet-block system lost during the collision process.

回答 (1)

Prof. Physics

2009-06-15 3:49 pm

✔ 最佳答案

a. Since there is no external force acting on the system, we use the law of conservation of momentum

Initial momentum = Final momentum

M2Vi = M1V + M2Vf

(0.050)(8.0 X 102) = (4.0)V + (0.050)[0.75(8.0 X 102)]

Speed of the block, V = 2.5 ms-1

Initial kinetic energy of the system

= 1/2 M2Vi2

= 1/2 (0.050)(8.0 X 102)2

= 16 000 J

Final kinetic energy of the system

= 1/2 M1V2 + 1/2 M2Vf2

= 1/2 (4.0)(2.5)2 + 1/2 (0.050)[0.75(8.0 X 102)]2

= 9012.5 J

Therefore, the mechanical energy lost by the system

= Initial K.E. - Final K.E.

= 16 000 - 9012.5

= 6987.5

= 6990 J

參考: Physics king

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