To solve this problem, let's address each part systematically.

1. Speed of the block/bullet combination after impact

The principle of conservation of momentum applies here:

$m_{\text{bullet}}v_{\text{bullet}} + m_{\text{block}}v_{\text{block}} = (m_{\text{bullet}} + m_{\text{block}})v_{\text{final}}$

Where:

• $m_{\text{bullet}} = 0.120 \, \text{kg}$
• $v_{\text{bullet}} = 400 \, \text{m/s}$
• $m_{\text{block}} = 2.5 \, \text{kg}$
• $v_{\text{block}} = 0 \, \text{m/s}$

Substitute values into the equation:

$(0.120)(400) + (2.5)(0) = (0.120 + 2.5)v_{\text{final}}$

$48 = 2.62v_{\text{final}}$

$v_{\text{final}} = \frac{48}{2.62} \approx 18.32 \, \text{m/s}$

So, the speed of the block/bullet combination is approximately 18.32 m/s.

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2. Magnitude and direction of the impulse on the bullet

Impulse ($J$) is the change in momentum, calculated as:

$J = \Delta p = m(v_{\text{final}} - v_{\text{initial}})$

For the bullet:

• Initial velocity $v_{\text{initial}} = 400 \, \text{m/s}$
• Final velocity $v_{\text{final}} = 18.32 \, \text{m/s}$

$J = (0.120)(18.32 - 400)$

$J = (0.120)(-381.68) \approx -45.8 \, \text{N·s}$

The impulse is 45.8 N·s, directed opposite to the initial motion (west).

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3. Magnitude and direction of the impulse on the block

By Newton's Third Law, the impulse exerted on the block by the bullet is equal in magnitude but opposite in direction to the impulse on the bullet.

Thus:

• Magnitude = 45.8 N·s
• Direction = east (same as the bullet's initial motion).

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4. Average force during the 5 ms interval

The average force ($F_{\text{avg}}$) is related to impulse by:

$F_{\text{avg}} = \frac{J}{\Delta t}$

Where:

• $J = 45.8 \, \text{N·s}$
• $\Delta t = 5 \, \text{ms} = 0.005 \, \text{s}$

$F_{\text{avg}} = \frac{45.8}{0.005} = 9160 \, \text{N}$

The average force is 9160 N.

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5. Percentage of kinetic energy remaining

Kinetic energy before impact ($KE_{\text{initial}}$):

$KE_{\text{initial}} = \frac{1}{2}m_{\text{bullet}}v_{\text{bullet}}^2$

$KE_{\text{initial}} = \frac{1}{2}(0.120)(400^2) = 9600 \, \text{J}$

Kinetic energy after impact ($KE_{\text{final}}$):

$KE_{\text{final}} = \frac{1}{2}(m_{\text{bullet}} + m_{\text{block}})v_{\text{final}}^2$

$KE_{\text{final}} = \frac{1}{2}(2.62)(18.32^2) \approx \frac{1}{2}(2.62)(335.56) \approx 439.8 \, \text{J}$

Percentage of kinetic energy remaining:

$\text{Percentage} = \left(\frac{KE_{\text{final}}}{KE_{\text{initial}}}\right) \times 100 = \left(\frac{439.8}{9600}\right) \times 100 \approx 4.58\%$

Approximately 4.58% of the initial kinetic energy remains.

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Final Answers:

1. Speed of the block/bullet combination: 18.32 m/s
2. Impulse on the bullet: 45.8 N·s (west)
3. Impulse on the block: 45.8 N·s (east)
4. Average force: 9160 N
5. Percentage of kinetic energy remaining: 4.58%

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Would you like detailed derivations or explanations for any of these answers?

Five Related Questions:

1. How would the results change if the block's mass were doubled?
2. What is the direction of the average force exerted on the block by the bullet?
3. How would friction on the table affect the calculation of $v_{\text{final}}$?
4. Can we calculate the distance the block/bullet travels after the collision?
5. What are some real-life implications of kinetic energy loss in collisions?

Tip:

Always check whether energy is conserved or lost in a system to understand how mechanical energy transforms during a process!