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+363m/sBlock 1Block 2(a) Before collision+0.677m/sblock 2mblock 2 16328mbullet 4.71gmblock 1 1244g(b) After collision


A 4.71-g bullet is moving horizontally with a velocity of +363 m/s, where the sign + indicates that it is moving to the right (see part a of the drawing). The bullet is approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part b. Note that both blocks are moving after the collision with the bullet. The mass of the first block is 1244 g, and its velocity is +0.677 m/s after the bullet passes through it. The mass of the second block is 1632 g. (a) What is the velocity of the second block after the bullet imbeds itself? (b) Find the ratio of the total kinetic energy after the collision to that before the collision.

Block 1
Block 2
(a) Before collision
block 2
mblock 2 16328
mbullet 4.71g
mblock 1 1244g
(b) After collision

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+363m/s Block 1 Block 2 (a) Before collision +0.677m/s block 2 mblock 2 16328 mbullet 4.71g mblock 1 1244g (b) After collision

Step 1


According to the law of conservation of momentum, total momentum of a system before the collision is equal to the total momentum of system after the collision.


mB is the mass of the bullet.

vB is the velocity of the bullet.

m1 is the mass of the first block.

m2 is the mass of the second block.

v1 is the velocity of the first block after the collision.

v2 is the velocity of the second block with bullet after the bullet imbeds itself.

Step 2

Rewrite the above equation for .

тзув — ту

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тзув — ту m,+M

Step 3

Substitute the ...

0.001kg(363m/s)-(1244g/0.001kg (0677m/s)
(1632g+4.71g 0.001kg

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0.001kg(363m/s)-(1244g/0.001kg (0677m/s) (4.71g) 1g (1632g+4.71g 0.001kg 1g (0.00471kg)(363m/s)-(1.244kg)(0.677m/s) (1.63671kg) -0530m/s


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