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A A 3.20 g bullet is fired horizontally at two blocks resting on a frictionless

ID: 2135908 • Letter: A

Question

A

A 3.20 g bullet is fired horizontally at two blocks resting on a frictionless tabletop as shown in Fig. 6-34a. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby imparted to the blocks as shown in Fig. 6-34b. Neglect the mass removed from the first block by the bullet.

A 3.20 g bullet is fired horizontally at two blocks resting on a frictionless tabletop as shown in Fig. 6-34a. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.630 m/s and 1.40 m/s, respectively, are thereby imparted to the blocks as shown in Fig. 6-34b. Neglect the mass removed from the first block by the bullet. Find the speed of the bullet immediately after it emerges from the first block. Find the bullet's original speed.

Explanation / Answer

well, moment before a collision equals total momentum ater a collison, so the bullet's initial momentum (mv) will equal the combined momentum of the blocks. There for, 3.3v= (1200*0.630)+(1800*1.40)
so, 3.3v = 756 + 2520
3.3v = 3276
v=992.73m/s < that's the answer to part (b)

To get (a) 3.3v = 2520
v = 840 m/s

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