A 4.90 g bullet is fired horizontally at two blocks resting on a smooth tabletop

Question

A 4.90 g bullet is fired horizontally at two blocks resting on a smooth tabletop, as shown in the top figure. 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 0.420 m/s and 1.06 m/s, respectively, are thereby imparted to the blocks, as shown in the bottom figure. Neglecting the mass removed from the first block by the bullet, find the speed of the bullet immediately after it emerges from the first block.

Explanation / Answer

assuming .42m/sec is in same direction of the velocity of bullet let initial velocity of bullet be v =>conservation of momentum we get .0049*v=1.2*.42+.0049*v1---------(1) =>conservation of momentum in second time we get .0049*v1=1.8049*1.06 =>v1=390.45m/sec sub v1 in above eqn 1 we get v=493.305 m/sec speed of the bullet immediately after it emerges from the first block.=v1=390.45m/sec

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.