Agohofidov6

2022-01-05

A 0.150-kg frame, when suspended from a coil spring, stretches the spring 0.070 m. A 0.200-kg lump of putty is dropped from rest onto the frame from a height of 30.0 cm. Find the maximum distance the frame moves downward from its initial position.

sirpsta3u

Beginner2022-01-06Added 42 answers

Step 1

The mass of the frame is ${m}_{A}=0.150kg$, the mass of the putty is ${m}_{B}=0.200kg$, the spring is initially stretched a distance ${x}_{o}=0.070m$ and the putty is dropped from rest from a height $h=0.030m$.

Step 2

We know that, in collision of all kinds, since there is no external net force on the system, the total momentum before the collision equals the total momentum after the collision:

$\overrightarrow{p}}_{1}={\overrightarrow{p}}_{2$

If the collision is between two particle (A and B), then we get:

$\overrightarrow{p}}_{A1}+{\overrightarrow{p}}_{B1}={\overrightarrow{p}}_{A2}+{\overrightarrow{p}}_{B2$

Where the momentum of a particle is given by:

$\overrightarrow{p}=m\overrightarrow{v}$

So we can write the collision equation in the following form:

$m}_{A}{\overrightarrow{v}}_{A1}+{m}_{B}{\overrightarrow{v}}_{B1}={m}_{A}{\overrightarrow{v}}_{A2}+{m}_{B}{\overrightarrow{v}}_{B2$ (1)

We also know that, when a spring is stretched (or compressed) a distance x, the restoring force is given by:

${F}_{x}=kx$

Where the elastic potential energy is given by:

$U}_{el}=\frac{1}{2}m{x}^{2$ (2)

Also, the gravitational potential energy is given by:

${U}_{grav}=mgy$ (3)

And the kinetic energy is given by:

$K=\frac{1}{2}m{v}^{2}$ (4)

Step 3

Calculations

We have three stages, the first stage is dropping the putty from its initial height until it reaches the frame, the second stage is the collision between the putty and the frame, and the third stage is the stretching of the spring to its maximum stretch point.

Step 4

The first stage:

When the putty is left to fall from its initial height, gravity is the only force that does work on it; So, from the conservation of energy, the gravitational potential energy at this height is totally converted into kinetic energy at the point where it meets the frame.

Therefore, by taking the zero potential energy level at the initial position of the frame and from equation (3) and (4), we get:

$m}_{B}gh=\frac{1}{2}{m}_{B}{v}_{B1}^{2$

$gh=\frac{1}{2}{v}_{B1}^{2}$

$v}_{B1}=\sqrt{2gh$

Now, we plug our value for h, so we get:

$v}_{B1}=\sqrt{2\times 9.8\times 0.300}=2.42\frac{m}{s$

This is the velocity of the putty just before the collision.

Step 5

The second stage:

When the putty reaches the frame. the net external force at the system is zero. and, at the moment of the collision, the spring is slightly stretched and we can ignore this change.

So, we can assume the conservation of momentum of the system during the collision.

Since the putty and the frame move together after the collision, they have the same velocity after the collision:

$v}_{A2}={v}_{B2}={v}_{2$

Now, we plug our values for $m}_{A},{m}_{B}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}{v}_{B1$ into equation (1), so we get:

$\left(0.150\right)\cdot \left(0\right)+\left(0.200\right)\cdot \left(2.42\right)=(0.150+0.200)\cdot {v}_{2}$

$v}_{2}=\frac{\left(0.200\right)\cdot \left(2.42\right)}{0.150+0.200}=1.38\frac{m}{s$

This is the velocity of the system right after the collision.

Step 6

The third stage:

First, we need to calculate the force constant k of the spring; So, before the collision, we apply Newton's second law to the frame along the vertical direction, so we get:

$\sum {F}_{y}=k{x}_{o}-{m}_{A}g=0$

abonirali59

Beginner2022-01-07Added 35 answers

The 0.15kg frame has a weight of 1.47N and this causes an extension of 0.07m.

This gives a spring constant of 21N/m

A 0.2kg lump of putty dropped from 0.30m will have

When the putty lands on the platform the spring will extend and this energy is stored in the spring.

The energy stored in a spring is 0.5F x ext or, since k(spring constant)

The spring will stop extending when the 0.589J of KE have been stored (absorbed) so

This gives e (extension) of 0.237m

This is the maximum extension and the spring will oscillate.

karton

Expert2022-01-11Added 613 answers

The equation for force

$F=kx\phantom{\rule{0ex}{0ex}}=mg\phantom{\rule{0ex}{0ex}}k=\frac{mg}{x}\phantom{\rule{0ex}{0ex}}=\frac{0.150kg\times 9.8\frac{m}{{s}^{2}}}{0.050m}\phantom{\rule{0ex}{0ex}}=29.4\frac{N}{m}$

When the total energy is conserved then equation for velocity is

$mgh=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}{v}^{2}=2gh\phantom{\rule{0ex}{0ex}}v=\sqrt{2gh}\phantom{\rule{0ex}{0ex}}=\sqrt{2\times 9.8\times 0.3}\phantom{\rule{0ex}{0ex}}=2.42\frac{m}{s}$

Now by using law of conservation of momentum

${m}_{1}{u}_{1}+{m}_{2}{u}_{2}={m}_{1}{v}_{1}+{m}_{2}{v}_{2}$

So the equation becomes

$mv+M\left(0\right)=mv+Mv\phantom{\rule{0ex}{0ex}}0.2\times 2.42+0=(0.15+0.2)v$

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