Question

A $500g$cart is released from rest $1.00m$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

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

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:

$\begin{array}{rr}{F}_{ramp}& =m{a}_{ramp}\\ mg\sin \theta & =m{a}_{ramp}\\ \Rightarrow a_{ramp}=9.81\text{m}/{\text{s}}^2\cdot 0.5& =4.9\text{m}/{\text{s}}^2\end{array}$

Then we can use equation for free fall:

$v_{initial}=\sqrt{2a_{ramp}{h}_{initial}}=\sqrt{2\cdot 4.9\text{m}/{\text{s}}^2\cdot 1\text{m}}=-3.13\text{m}/\text{s}$

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

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

Finally :

$h_{final}=\frac{v_{\text{f}}^2}{2a_{ramp}}=\frac{4.8841{\text{m}}^2/{\text{s}}^2}{2\cdot 4.9\text{m}/{\text{s}}^2}=0.498\text{m}$