Ch 11: Impulse and Momentum

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook

Knight Calc 5th Edition

Ch 11: Impulse and Momentum

Problem 49b

Chapter 11, Problem 49b

What is the speed of a 10g bullet that, when fired into a 10kg stationary wood block, causes the block to slide 5.0 cm across a wood table?

Verified step by step guidance

Convert all given quantities to SI units. The mass of the bullet is 10 g, which is equivalent to 0.01 kg. The distance the block slides is 5.0 cm, which is equivalent to 0.05 m.

Apply the principle of conservation of momentum to the collision. Since the block is initially stationary, the total momentum before the collision is the momentum of the bullet: \( p_{initial} = m_{bullet} \cdot v_{bullet} \). After the collision, the bullet and block move together, so \( p_{final} = (m_{bullet} + m_{block}) \cdot v_{final} \). Set \( p_{initial} = p_{final} \) to relate \( v_{bullet} \) and \( v_{final} \).

Determine the velocity of the block and bullet system after the collision (\( v_{final} \)) using the work-energy principle. The block slides a distance of 0.05 m against the frictional force. The work done by friction equals the kinetic energy of the block and bullet system: \( f_{friction} \cdot d = \frac{1}{2} (m_{bullet} + m_{block}) \cdot v_{final}^2 \).

Calculate the frictional force \( f_{friction} \) using \( f_{friction} = \mu \cdot m_{block} \cdot g \), where \( \mu \) is the coefficient of kinetic friction between the block and the table, and \( g \) is the acceleration due to gravity (9.8 m/s²). If \( \mu \) is not provided, it must be assumed or given in the problem.

Combine the equations from the conservation of momentum and the work-energy principle to solve for the initial speed of the bullet \( v_{bullet} \). Substitute known values and solve algebraically, ensuring units are consistent throughout the calculation.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event. In this scenario, the bullet and the block form a system where the bullet's momentum is transferred to the block upon impact, allowing us to calculate the bullet's speed based on the resulting motion of the block.

Kinetic Energy and Work-Energy Principle

The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. When the bullet embeds itself in the block, it transfers kinetic energy to the block, which then slides across the table. The distance the block slides can be used to determine the initial kinetic energy imparted by the bullet.

Friction and Deceleration

Friction is the force that opposes the motion of an object sliding over a surface. In this case, the block experiences friction as it slides across the wood table, which decelerates it. Understanding the coefficient of friction between the block and the table is essential for calculating how far the block will slide after being struck by the bullet.

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Related Practice

Textbook Question

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is μ_k .Find an expression for the bullet's speed v_bullet.

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Textbook Question

Two 500 g blocks of wood are 2.0 m apart on a frictionless table. A 10 g bullet is fired at 400 m/s toward the blocks. It passes all the way through the first block, then embeds itself in the second block. The speed of the first block immediately afterward is 6.0 m/s . What is the speed of the second block after the bullet stops in it?

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Textbook Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides 8.2 m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction.

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Textbook Question

A 1500 kg weather rocket accelerates upward at 10 m/s². It explodes 2.0 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion?

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Textbook Question

Most geologists believe that the dinosaurs became extinct 65 million years ago when a large comet or asteroid struck the earth, throwing up so much dust that the sun was blocked out for a period of many months. Suppose an asteroid with a diameter of 2.0 km and a mass of 1.0×10¹³ kg hits the earth (6.0×10²⁴ kg) with an impact speed of 4.0×10⁴ m/s. What percentage is this of the earth's speed around the sun? The earth orbits the sun at a distance of 1.5×10¹¹ m .

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