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Chapter 7: Problem 70

A sled of mass \(5.0 \mathrm{kg}\) is coasting along on a frictionless ice- covered lake at a constant speed of \(1.0 \mathrm{m} / \mathrm{s} .\) A \(1.0-\mathrm{kg}\) book is dropped vertically onto the sled. At what speed does the sled move once the book is on it?

Short Answer

Expert verified
Answer: 0.833 m/s

Step by step solution

01

Write the conservation of linear momentum equation

To solve this problem, we need to use the conservation of linear momentum equation which states that the total momentum before an event is equal to the total momentum after that event. Mathematically, we can write this as: \( m_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_f \) Here, \(m_{1}\) is the mass of the sled, \(v_{1}\) is its initial velocity, \(m_{2}\) is the mass of the book, and \(v_{2}\) is the velocity of the book when it's dropped (which is 0, as it's dropped vertically). We need to find the final velocity \(v_f\) of the sled and the book when they move together.
02

Calculate the initial momentum of the sled

First, we need to calculate the initial momentum of the sled before the book is dropped onto it. To do this, multiply the mass of the sled by its initial velocity: \( p_{1} = m_{1}v_{1} = 5.0 \text{ kg} \times 1.0 \text{ m/s} = 5.0 \text{ kg m/s} \)
03

Calculate the initial momentum of the book

As the book is dropped vertically, its horizontal velocity is zero. Thus, the initial momentum of the book is: \( p_{2} = m_{2}v_{2} = 1.0 \text{ kg} \times 0 \text{ m/s} = 0 \text{ kg m/s} \)
04

Substitute values into the conservation of linear momentum equation

Now, we have all the values and can use the conservation of linear momentum equation to find the final velocity of the sled and the book: \( 5.0 \text{ kg m/s} + 0 \text{ kg m/s} = (5.0 \text{ kg} + 1.0 \text{ kg})v_f \)
05

Solve for the final velocity

Now, solve the equation to find the final velocity of the sled and the book: \( 5.0 \text{ kg m/s} = 6.0 \text{ kg} \cdot v_f \) Then, divide both sides by 6.0 kg to get: \( v_f = \frac{5.0 \text{ kg m/s}}{6.0 \text{ kg}} = 0.833 \text{ m/s} \) Hence, once the book is on the sled, they move together at a speed of \(0.833 \text{ m/s}\).

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Most popular questions from this chapter

An automobile traveling at a speed of \(30.0 \mathrm{m} / \mathrm{s}\) applies its brakes and comes to a stop in 5.0 s. If the automobile has a mass of $1.0 \times 10^{3} \mathrm{kg},$ what is the average horizontal force exerted on it during braking? Assume the road is level.
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