A 5.0-gram bullet is shot vertically into a 1.2-kg block. The bullet penetrates the block and becomes embedded in the block, lifting the block upward by 6.0 cm (see the figure below). Assuming that air resistance is negligible, the initial speed of the bullet is closest to 单项选择题
A
140 m/s
B
360 m/s
C
180 m/s
D
260 m/s
E
120 m/s

Question

A 5.0-gram bullet is shot vertically into a 1.2-kg block. The bullet penetrates the block and becomes embedded in the block, lifting the block upward by 6.0 cm (see the figure below). Assuming that air resistance is negligible, the initial speed of the bullet is closest to 单项选择题

A

140 m/s

B

360 m/s

C

180 m/s

D

260 m/s

E

120 m/s

BlackTom AI · Accepted Answer

Problem restatement: A 5.0 g bullet is shot into a 1.2 kg block and embeds in it. After embedding, the combined mass rises 6.0 cm. Air resistance is negligible. We are asked to find the initial speed of the bullet.

Step 1: Determine the post-collision velocity of the combined mass. Since the bullet embeds in the block, the collision is perfectly inelastic and momentum is conserved. The total mass after the collision is M = m_b + m_block = 0.005 kg + 1.2 kg = 1.205 kg.

Step 2: Use energy conservation to relate post-collision velocity to the height reached. The embedded system climbs and its kinetic energy is converted to gravitational potential energy: (1/2) M v_post^2 = M g h, so v_post^2 = 2 g h. Here h = 0.06 m and g ≈ 9.8 m/s^2. Therefore v_post = sqrt(2 × 9.8 × 0.06) ≈ sqrt(1.176) ≈ 1.08 m/s.

Step 3: Use momentum conservation to relate the bullet’s initial speed to the post-collision speed. Before the collision, the block is at rest and the bullet has speed v_b; after the collision, the bullet and block move together at v_post. Momentum conservation gives m_b v_b = M v_post, so v_b = M v_post / m_b. Substituting, v_b ≈ (1.205 kg × 1.08 m/s) / 0.005 kg ≈ (1.3034) / 0.005 ≈ 260.7 m/s, i.e., about 2.6 × 10^2 m/s.

Option analysis:
- 140 m/s: If the bullet’s speed were only ~140 m/s, the post-collision velocity v_post = (m_b v_b)/M would be far smaller than the 1.08 m/s required to rise 0.06 m. This would yield a much smaller height than 0.06 m, contradicting the given rise. Therefore this option is inconsistent with the energy and momentum balances.
- 360 m/s: A higher initial speed would produce a proportionally higher post-collision velocity, leading to a larger height than 0.06 m when converted to gravitational potential. Since the observed rise is only 0.06 m, this overestimates the needed v_b, making this option unlikely.
- 180 m/s: This would give a post-collision velocity v_post of about (m_b v_b)/M ≈ (0.005 × 180)/1.205 ≈ 0.75 m/s, which would correspond to a height of h ≈ v_post^2/(2g) ≈ (0.75^2)/(2×9.8) ≈ 0.028 m, significantly less than 0.06 m. So this speed is too low.
- 260 m/s: This value, when processed through the momentum and energy relations, yields v_post ≈ 1.08 m/s and a maximum height ≈ 0.06 m, matching the given data. It is consistent with both momentum conservation during embedding and energy conservation during ascent.
- 120 m/s: This would produce an even smaller v_post (≈ (0.005 × 120)/1.205 ≈ 0.498 m/s), giving a height of h ≈ (0.498^2)/(2×9.8) ≈ 0.0127 m, far short of the observed 0.06 m. Hence this option is not feasible.

In summary, evaluating the implications of each candidate initial speed against both momentum conservation (for the embedding event) and energy conservation (for the ascent) shows that 260 m/s is the consistent choice with the observed 6.0 cm rise.

A 5.0-gram bullet is shot vertically into a 1.2-kg block. The bullet penetrates the block and becomes embedded in the block, lifting the block upward by 6.0 cm (see the figure below). Assuming that air resistance is negligible, the initial speed of the bullet is closest to 单项选择题

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