Chapter 11: Q. 40 (page 289)

A $500 g$ cart is released from rest $1.00 m$ from the bottom of a frictionless, $30.0 °$ ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. FIGURE P $11.40$ shows the force during the collision. After the cart bounces, how far does it roll back up the ramp?

Short Answer

Expert verified

It roll back up the ramp is $0.498 m$ .

Step by step solution

Given Information

We need to find far from it roll back up the ramp.

Simplify

In order to find velocity if cart at the bottom of frictionless ramp (and to use equation for free fall), we have to find acceleration down the frictionless ramp:

$\begin{array}{r} F_{r a m p} & = m a_{r a m p} \\ m g \sin θ & = m a_{r a m p} \\ \Rightarrow a_{r a m p} = 9.81 m / s^{2} \cdot 0.5 & = 4.9 m / s^{2} \end{array}$

Then we can use equation for free fall:

$v_{i n i t i a l} = \sqrt{2 a_{r a m p} h_{i n i t i a l}} = \sqrt{2 \cdot 4.9 m / s^{2} \cdot 1 m} = - 3.13 m / s$

Here minus sign indicates the cart is going down.
From impulse-momentum theorem we have:

$\begin{array}{r} p_{f} & = p_{i} + J \\ m v_{f} & = m v_{i} + J \\ 0.5 kg \cdot v_{f} & = 0.5 kg \cdot (- 3.13) m / s + \frac{1}{2} \cdot 200 N \cdot 26.7 \cdot 10^{- 3} s \\ v_{f} & = 2.21 m / s. \end{array}$

Finally :

$h_{f i n a l} = \frac{v_{f}^{2}}{2 a_{r a m p}} = \frac{4.8841 m^{2} / s^{2}}{2 \cdot 4.9 m / s^{2}} = 0.498 m$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

A $500 g$ cart is released from rest $1.00 m$ from the bottom of a frictionless, $30.0 °$ ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. FIGURE P $11.40$ shows the force during the collision. After the cart bounces, how far does it roll back up the ramp?

Short Answer

Step by step solution

Given Information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Classical Mechanics

Energy Physics

Radiation

Measurements

Fields in Physics

Study anywhere. Anytime. Across all devices.

A 500gcart is released from rest 1.00mfrom the bottom of a frictionless, 30.0°ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. FIGURE P11.40shows the force during the collision. After the cart bounces, how far does it roll back up the ramp?

Short Answer

Step by step solution

Given Information

Simplify

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Classical Mechanics

Energy Physics

Radiation

Measurements

Fields in Physics

Study anywhere. Anytime. Across all devices.

A $500 g$ cart is released from rest $1.00 m$ from the bottom of a frictionless, $30.0 °$ ramp. The cart rolls down the ramp and bounces off a rubber block at the bottom. FIGURE P $11.40$ shows the force during the collision. After the cart bounces, how far does it roll back up the ramp?