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3 years ago

5

A 5.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 8.0 cm if the marble is to just reac

h a target 20 m above the marble’s position on the compressed spring. (a) What is the change ΔUg in the gravitational potential energy of the marble–Earth system during the 20 m ascent? (b) What is the change ΔUs in the elastic potential energy of the spring during its launch of the marble? (c) What is the spring constant of the spring?

1 answer:

taurus [48]3 years ago

5 0

Answer:

a. 0.98 J

b. -0.98 J

c. 306.25 J/m²

Explanation:

(a)

$\Delta U_g = mgh\\\Delta U_g = 5\times 10^{-3}\times 9.8 \times 20\\\Delta U_g = 0.980 J$

(b)

The increase in gravitational potential energy = decrease in elastic potential energy

Elastic potential energy = -0.980 J

(c) The spring constant can be calculated as follows:

$-\frac{1}{2}kx^2=-0.980\\ k = \frac{2\times 0.980}{0.08^2} = 306.25 J/m^2$

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Suppose a baseball pitcher throws the ball to his catcher.

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a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

$\Delta v = |v-u|$

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

$\Delta v=|v-0|=|v|$

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

$\Delta v =|0-v|=|-v|$

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

$\Delta p = m \Delta v$

where

m is the mass of the object

$\Delta v$ is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

$|\Delta p|=m|\Delta v|$

- In situation 1 (pitcher throwing the ball), the change in momentum is

$\Delta p = m|\Delta v|=m|v|=mv$

- In situation 2 (catcher receiving the ball), the change in momentum is

$\Delta p = m\Delta v = m|-v|=mv$

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

$I=\Delta p$

where

I is the impulse

$\Delta p$ is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

$I=F\Delta t$

where

I is the impulse

F is the force applied

$\Delta t$ is the time of impact

This can be rewritten as

$F=\frac{I}{\Delta t}$

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- $\Delta t$ when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to $\Delta t$ , this means that the force is larger when the ball is catched.

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Answer:

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c is the speed of light

Here we know

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L = 35.0 m

So we can re-arrange the equation to find the length of the rocket at rest:

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4 years ago

It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp

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Explanation:

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So,

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