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

Question

A 3.20 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 0.450 m/s and 1.00 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. Find the bullet's original speed.

Explanation / Answer

mass of bullet mb = 3.2 g

mass of block1 = m1 = 1.2 kg

velocity of block1 v1 = 0.45 m/s

mass of block2 = m2 = 1.8 kg

velocity of block2 = 1 m/s

by conservation of momentum

mb * vb = m1 * v1 + (m2 + mb) * v2

3.2 * 10^-3 * vb = 1.2 * 0.45 + (3.2 * 10^-3 + 1.8) * 1

vb = 732.25 m/s

the bullet's original speed = 732.25 m/s

momentum of the first block = 1.2 * 0.45

momentum of the first block = 0.54 kgm/s

momentum of the bullet = 3.2 * 10^-3 * 732.25

momentum of the bullet = 2.3432 kgm/s

momentum left = 2.3432 - 0.54

momentum left = 1.8032 kgm/s

3.2 * 10^-3 * vf = 1.8032

vf = 563.5 m/s

speed of the bullet immediately after it emerges from the first block = 563.5 m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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