New Notes Blog: A 100g cart is placed against a spring with a force constant of 20,000 N/m. The spring is compressed 2.5cm. A second cart with a mass of 200g is...

Saturday, 14 February 2015

A 100g cart is placed against a spring with a force constant of 20,000 N/m. The spring is compressed 2.5cm. A second cart with a mass of 200g is...

Hello!

Denote the mass of the first cart as $\displaystyle{m}_{{1}},$ the mass of the second cart as $\displaystyle{m}_{{2}},$ the spring's force constant as $\displaystyle{k}_{{s}}$ and the coefficient of friction as $\displaystyle{k}_{{f{.}}}$ Denote the distance of spring compression as $\displaystyle{x}.$

The friction force will be $\displaystyle{0.25}\cdot{m}_{{1}}\cdot{g{\approx}}{0.25}{N}.$ The starting spring force is $\displaystyle{20000}\frac{{N}}{{m}}\cdot{0.025}{m}={500}{N}.$ Therefore we can neglect the friction force while spring is releasing. Also we can neglect the position change while spring is releasing...

Hello!

Determine the speed V_0 of the first cart after the spring is released completely. Use the law of energy conservation: the elastic potential energy is $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot{k}_{{s}}\cdot{{x}}^{{2}}$ transforms into kinetic energy of the first car, $\displaystyle{\left(\frac{{1}}{{2}}\right)}\cdot{m}_{{1}}\cdot{{V}_{{0}}^{{2}}}.$

So $\displaystyle{V}_{{0}}={x}\cdot\sqrt{{\frac{{k}_{{s}}}{{m}_{{1}}}}}={0.025}\cdot\sqrt{{{200},{000}}}\approx{11.2}{\left(\frac{{m}}{{s}}\right)}.$

Then the first cart decelerates uniformly due to the friction force $\displaystyle{k}_{{f{\cdot}}}{m}_{{1}}\cdot{g{.}}$ It gives the negative acceleration of $\displaystyle{a}={k}_{{f{\cdot}}}{g{\approx}}{2.45}{\left(\frac{{m}}{{{s}}^{{2}}}\right)}$ (Newton's Second law).

The speed is $\displaystyle{V}{\left({t}\right)}={V}_{{0}}-{a}\cdot{t},$ the position is $\displaystyle{L}{\left({t}\right)}={V}_{{0}}\cdot{t}-{a}\cdot\frac{{{t}}^{{2}}}{{2}}.$

Let's find the (smallest) time $\displaystyle{t}_{{1}}$ when $\displaystyle{L}{\left({t}_{{1}}\right)}$ becomes 5m (and collision happens):

$\displaystyle{2.45}{{t}_{{1}}^{{2}}}-{22.4}{t}_{{1}}+{10}={0},$

$\displaystyle{t}_{{1}}=\frac{{{22.4}-\sqrt{{{{22.4}}^{{2}}-{4}\cdot{2.45}\cdot{10}}}}}{{4.9}}\approx{0.47}{\left({s}\right)}.$

The speed before collision will be $\displaystyle{V}_{{1}}={V}_{{0}}-{a}\cdot{t}_{{1}}\approx{11.2}-{2.45}\cdot{0.47}\approx{10}{\left(\frac{{m}}{{s}}\right)}.$

Now for the collision. Elastic collision means no bounce, both bodies move as a whole. Then the law of impulse conservation gives that the final speed $\displaystyle{V}_{{2}}$ satisfies

$\displaystyle{m}_{{1}}\cdot{V}_{{1}}={\left({m}_{{1}}+{m}_{{2}}\right)}\cdot{V}_{{2}},$ or

$\displaystyle{V}_{{2}}={\left(\frac{{m}_{{1}}}{{{m}_{{1}}+{m}_{{2}}}}\right)}\cdot{V}_{{1}}=\frac{{V}_{{1}}}{{3}}\approx{3.33}{\left(\frac{{m}}{{s}}\right)}.$ This is the answer.

New Notes Blog

Saturday, 14 February 2015

A 100g cart is placed against a spring with a force constant of 20,000 N/m. The spring is compressed 2.5cm. A second cart with a mass of 200g is...

No comments:

Post a Comment

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Saturday, 14 February 2015

A 100g cart is placed against a spring with a force constant of 20,000 N/m. The spring is compressed 2.5cm. A second cart with a mass of 200g is...

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?