Question

A 10g bullet moving at 200m/s stops after penetrating 5cm of wooden plank.The average force exerted on block will be?

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Answer 1

Explanation:

350m/s

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Answer 2

Answer:

• -4000 N.

Explanation:

Given:

• Mass (m) = 10g
• Initial velocity (u) = 200 m/s
• Final velocity (v) = 0 m/s
• Displacement (s) = 5 cm

To find:

• Average force (F)

Solution:

Final velocity is 0 since the bullet comes at rest at a particular point of time.

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Conversions:

=> 10g = 10/1000 Kg

=> Mass (m) = 0.01 Kg

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=> 5 cm = 5/100 m

=> Displacement (s) = 0.05 m

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Using Kinematics:

• v² - u² = 2as

=> 0²- (200)² = 2a x 0.05

=> 0 - 40000 = 2a × 0.05

=> -40000 = 0.1a

=> a = -40000/0.1

∴ a = -400000

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F = Ma

=> F = 0.01(-400000)

∴ F = -4000 N

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∴ We can say that the force is acting in the opposite direction, retarding force.