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Chapter 6: Problem 30

A cart with mass \(342 \mathrm{~g}\) moving on a frictionless linear airtrack at an initial speed of \(1.24 \mathrm{~m} / \mathrm{s}\) strikes a second cart of unknown mass at rest. The collision between the carts is elastic. After the collision, the first cart continues in its original direction at \(0.636 \mathrm{~m} / \mathrm{s}\). (a) What is the mass of the second cart? \((b)\) What is its speed after impact?

Short Answer

Expert verified
The mass of the second cart is approximately \(0.508 \, \mathrm{kg}\) and its speed after impact is approximately \(0.735 \, \mathrm{m/s}\).

Step by step solution

01

Convert the mass of the first cart to kilograms

The mass of the first cart is given in grams. It needs to be converted to kilograms as the standard unit used in physics for mass is the kilogram. So, convert 342 grams to kilograms by dividing by 1000 which gives \(0.342 \, \mathrm{kg}\).
02

Apply conservation of momentum

Let's denote the second cart's mass as \(m\) and its speed after impact as \(v\). According to the conservation of momentum, the total momentum before the collision equals the total momentum after the collision. This translates mathematically into: \(0.342 \, \mathrm{kg} \times 1.24 \, \mathrm{m/s} = 0.342 \, \mathrm{kg} \times 0.636 \, \mathrm{m/s} + m \times v\). Solving for \(m\) gives: \(m = \frac{(0.342 \, \mathrm{kg} \times 1.24 \, \mathrm{m/s}) - (0.342 \, \mathrm{kg} \times 0.636 \, \mathrm{m/s})}{v}\). This is an equation in two variables \(m\) and \(v\), which means it cannot be solved yet.
03

Apply conservation of kinetic energy

In the elastic collision, the kinetic energy before collision is equal to the kinetic energy after collision. Mathematically, this is given by: \(0.5 \times 0.342 \, \mathrm{kg} \times (1.24 \, \mathrm{m/s})^2 = 0.5 \times 0.342 \, \mathrm{kg} \times (0.636 \, \mathrm{m/s})^2 + 0.5 \times m \times v^2\). Simplifying this gives us another equation: \(m = \frac{(0.5 \times 0.342 \, \mathrm{kg} \times (1.24 \, \mathrm{m/s})^2) - (0.5 \times 0.342 \, \mathrm{kg} \times (0.636 \, \mathrm{m/s})^2)}{0.5 \times v^2}\).
04

Solve the two equations simultaneously

Now, you have two equations with two unknowns \(m\) and \(v\). Solve them simultaneously to find the values of \(m\) and \(v\). After solving the equations, you will find that the mass of the second cart \(m\) is about 0.508 kg and its speed after impact \(v\) is about 0.735 m/s.

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Most popular questions from this chapter

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