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3 years ago

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A 0.032-kg bullet is fired vertically at 230 m/s into a 0.15-kg baseball that is initially at rest. How high does the combined b

ullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?

1 answer:

gogolik [260]3 years ago

5 0

Answer:

$heigth=83.44m$

Explanation:

Given data

Baseball mass m₁=0.15 kg

initial speed v₁=0

Bullet mass m₂=0.032 kg

final speed v₂=230 m/s

To find

height h=?

Solution

From conservation of momentum we know that

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (0.032kg)(230m/s)+(0.15kg)(0m/s)=(0.15kg+0.032kg)v\\7.36+0=0.182v\\v=7.36/0.182\\v=40.44m/s$

Now from the conservation of mechanical energy

$P.E=K.E\\mgh=(1/2)mv^{2}\\ gh=(1/2)v^{2}\\ (9.8m/s^{2} )h=(1/2)(40.44m/s)^{2}\\ (9.8m/s^{2} )h=(817.7m^{2} /s^{2} )\\h=(817.7m^{2} /s^{2} )/9.8m/s^{2}\\ heigth=83.44m$

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A car is traveling at 7.0 m/s when the driver applies the brakes. The car moves 1.5 m before it comes to a complete stop. If the

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Answer:

d. 6.0 m

Explanation:

Given;

initial velocity of the car, u = 7.0 m/s

distance traveled by the car, d = 1.5 m

Assuming the car to be decelerating at a constant rate when the brakes were applied;

v² = u² + 2(-a)s

v² = u² - 2as

where;

v is the final velocity of the car when it stops

0 = u² - 2as

2as = u²

a = u² / 2s

a = (7)² / (2 x 1.5)

a = 16.333 m/s

When the velocity is 14 m/s

v² = u² - 2as

0 = u² - 2as

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The correct option is d

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2 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

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Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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Answer:

a. The thickness of the wire is 2.5 mm.

b. The wire is 0.25 cm thick.

Explanation:

Number of turns of the wire = 10

The length of total turns = 25 mm

a. The thickness of the wire can be determined by;

thickness of the wire = $\frac{length of total turns}{number of turns}$

= $\frac{25}{10}$

= 2.5 mm

Therefore, the wire is 2.5 mm thick.

b. To determine the thickness of the wire in centimetre;

10 mm = 1 cm

So that,

2.5 mm = x

x = $\frac{2.5}{10}$

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The wire is 0.25 cm thick.

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