(hrwc9p49) A 6.15 g bullet is fired horizontally at two blocks resting on a smoo
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Question
(hrwc9p49) A 6.15 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 1.02 m/s, respectively, are thereby imparted to the blocks, as shown in the bottom figure.
A. 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.
B. Find the bullet's original speed.
(hrwc9p49) A 6.15 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 1.02 m/s, respectively, are thereby imparted to the blocks, as shown in the bottom figure. A. 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. B. Find the bullet's original speed.Explanation / Answer
v1_bullet: bullet's speed before striking either block
v2_bullet: bullet's speed after exiting first block
v_block1: speed of block 1 after bullet exits (0.42 m/s)
v_block2: speed of block 2 after bullet embeds (1.02 m)--this is also the bullet's final speed
m_bullet: mass of bullet (0.00615 kg)
m_block1: mass of block 1 (1.2 kg)
m_block2: mass of block 2 (1.8 kg)
Total momentum before initial strike:
(m_bullet)(v1_bullet)
Total momentum after bullet exits block 1:
(m_bullet)(v2_bullet) + (m_block1)(v_block1)
Total momentum after bullet embeds in block2:
(m_block1)(v_block1) + (m_bullet+m_block2)(v_block2)
The conservation law says that all three of these are equal:
(m_bullet)(v1_bullet) = (m_bullet)(v2_bullet) + (m_block1)(v_block1)
(m_bullet)(v1_bullet) = (m_block1)(v_block1) + (m_bullet+m_block2)(v_block2)
You can rewrite the above as two separate equations:
(m_bullet)(v1_bullet) = (m_bullet)(v2_bullet) + (m_block1)(v_block1) ............(1)
(m_bullet)(v1_bullet) = (m_block1)(v_block1) + (m_bullet+m_block2)(v_block2)..............(2)
Solving (2)
0.00615 * (v1_bullet) = 1.2*0.42 + (1.2 + 1.8) * 1.02
0.00615 * (v1_bullet) = 3.565
(v1_bullet) = 579.67 m/s....................part (b)
Solving (1)
(m_bullet)(v1_bullet) = (m_bullet)(v2_bullet) + (m_block1)(v_block1)
0.00615 * 579.67 = 0.00615 * (v2_bullet) + 1.2 * 0.42
3.565 = 0.00615 * (v2_bullet) + 0.504
3.06 = 0.00615 * (v2_bullet)
(v2_bullet) = 497.56 m/s.............part (a)
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