Chapter 11: Q 78 (page 291)

$(0.10 k g + 0.20 k g) v_{1 x} = (0.10 k g) (3.0 m / s) \frac{1}{2} (0.30 k g) (0 m / s^{2}) + \frac{1}{2} (3.0 N / m) {(∆ x_{2})}^{2} = \frac{1}{2} (0.30 k g) {(v_{1 x})}^{2} + \frac{1}{2} (3.0 N / m) {(0 m)}^{2}$
a. Write a realistic problem for which this is the correct equation(s).
b. Finish the solution of the problem, including a pictorial representation.

Short Answer

Expert verified

a) The realistic problem will be "100g ball moves with speed 3.0 m/s collides with 200g ball at rest and move together towards an attached spring 3.0 N/m to a wall. Find the displacement of the spring after the collision of the combination with it"

b) Displacement, $∆ x_{2} = 0.18 m$

Step by step solution

Step 1. Given equation is :0.10 kg+0.20 kgv1x=(0.10 kg)(3.0 m/s)12(0.30 kg)(0 m/s2)+12(3.0 N/m)(∆x2)2 =12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2

We need to find out a realistic problem for which this is the correct equation and have to find out the solution including pictorial representation.

Step 2. Expressing the realistic problem of the given equation

The realistic problem will be :

"100g ball moves with speed 3.0 m/s collides with 200g ball at rest and move together towards an attached spring 3.0 N/m to a wall. Find the displacement of the spring after the collision of the combination with it"

Step 3. Solving the equations to find the speed of the combination and the displacement.

$S o l v i n g t h f i r s t e q u a t i o n, (0.10 k g + 0.20 k g) v_{1 x} = (0.10 k g) (3.0 m / s) v_{1 x} = \frac{0.30 k g m / s}{0.30 k g} = 1 m / s S o l v i n g s e c o n d e q u a t i o n f o r ∆ x_{2}, \frac{1}{2} (0.30 k g) (0 m / s^{2}) + \frac{1}{2} (3.0 N / m) {(∆ x_{2})}^{2} = \frac{1}{2} (0.30 k g) {(v_{1 x})}^{2} + \frac{1}{2} (3.0 N / m) {(0 m)}^{2} \frac{1}{2} (3.0 N / m) {(∆ x_{2})}^{2} = \frac{1}{2} (0.30 k g) {(v_{1 x})}^{2} {(∆ x_{2})}^{2} = 0.10 {(v_{1 x})}^{2} {(∆ x_{2})}^{2} = 0.10 {(1.0)}^{2} ∆ x_{2} = 0.18 m$

Pictorial Representation,

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Short Answer

Step by step solution

Step 1. Given equation is :0.10 kg+0.20 kgv1x=(0.10 kg)(3.0 m/s)12(0.30 kg)(0 m/s2)+12(3.0 N/m)(∆x2)2 =12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2

Step 2. Expressing the realistic problem of the given equation

Step 3. Solving the equations to find the speed of the combination and the displacement.

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0.10kg+0.20kgv1x=(0.10kg)(3.0m/s)12(0.30kg)(0m/s2)+12(3.0N/m)(∆x2)2=12(0.30kg)(v1x)2+12(3.0N/m)(0m)2a. Write a realistic problem for which this is the correct equation(s).b. Finish the solution of the problem, including a pictorial representation.

Short Answer

Step by step solution

Step 1. Given equation is :0.10 kg+0.20 kgv1x=(0.10 kg)(3.0 m/s)12(0.30 kg)(0 m/s2)+12(3.0 N/m)(∆x2)2 =12(0.30 kg)(v1x)2+12(3.0 N/m)(0 m)2

Step 2. Expressing the realistic problem of the given equation

Step 3. Solving the equations to find the speed of the combination and the displacement.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physics of Motion

Measurements

Modern Physics

Force

Translational Dynamics

Rotational Dynamics

Study anywhere. Anytime. Across all devices.