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luda_lava [24]
3 years ago
5

a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric

tionless surface, and is thin enough that the bullet passes completely through it. Immediately after the bullet exits the block, the speed of the block is 23m/s. (a) what is the speed of the when it exits the block? (b) is the final kinetic energy of this system equal to, less than, or greater than the initial kinetic energy? Explain. (c) verify your answer to part (b) by calculating the initial and final kinetic energies of the system.
Physics
1 answer:
n200080 [17]3 years ago
6 0

Part a)

Here in this we can use momentum conservation as there is no external force on it

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}

here we know that

m_1 = 0.004 kg

v_{1i} = 650 m/s

m_2 = 0.095

v_{2i} = 0

v_{2f} = 23 m/s

now by above equation

0.004*650 + 0.095* 0 = 0.004*v + 0.095*23

2.6 + 0 = 0.004*v + 2.185

v = 103.75 m/s

Part b)

Final kinetic energy of the system

KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2

KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2

KE_f = 46.65 J

Initial Kinetic energy of the system will be

KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2

KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2

KE_f = 845 J

So here kinetic energy is decreased for this system

final energy is less than initial energy

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A basketball center holds a basketball straight out, 2.0 m above the floor, and releases it. It bounces off the floor and rises
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Answer:

a) The velocity of the ball before it hits the floor is -6.3 m/s

b) The velocity of the ball after it hits the floor is 3.1 m/s

c) The magnitude of the average acceleration is 470 m/s². The direction is upward at an angle of 90º with the ground.

Explanation:

First, let´s calcualte how much time it takes the ball to hit the floor:

The equation for the position of the ball is:

y = y0 + v0 * t + 1/2 g * t²

Where:

y = position at time t

y0 = initial position

v0 = initial velocity

t = time

g = acceleration due to gravity

We take the ground as the origin of the reference system.

a) Since the ball is realesed and not thrown, the initial velocity v0 is 0. The direction of the acceleration is downward, towards the origin, then "g" will be negative. When the ball hits the ground its position will be 0. Then:

0 = 2.0 m + 0 m/s *t - 1/2 * 9.8 m/s²  * t²

-2.0 m = -4.9 m/s²  * t²

t² = -2.0 m / - 4.9 m/s²

t = 0.64 s

The equation for the velocity of a falling object is:

v = v0 + g * t      where "v" is the velocity

since v0= 0:

v = g * t = -9.8 m/s² * 0.64 s = -6.3 m/s

b) Now, we know that the velocity of the ball when it reaches the max height must be 0. We can obtain the time it takes the ball to reach that height from the equation for velocity and then use that time in the equation for position to obtain the initial velocity:

v = v0 + g * t

0 = v0 + g * t

-v0/g = t

now we replace t in the equation for position, since we know that the maximum height is 1.5 m:

y = y0 + v0 * t + 1/2* g * t²           y = 1.5 m       y0 = 0 m   t = -v0/g

1.5 m = v0 * (-v0/g) + 1/2 * g (-v0/g)²

1.5 m = - v0²/g - 1/2 * v0²/g

1.5 m = -3/2 v0²/g

1.5 m * (-2/3) * g = v0²

1.5 m * (-2/3) * (-9.8 m/s²) = v0²

v0 = 3.1 m/s

c) The average acceleration will be:

a = final velocity - initial velocity / time

a = 3.1 m/s - (-6.3 m/s) / 0.02 s = 470 m/s²

the direction of the acceleration is upward perpendicular to the ground.

The vector average acceleration will be:

a = (0, 470 m/s²) or (470 m/s² * cos 90º, 470 m/s² * sin 90º)

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WITCHER [35]

Answer:

The angle of refraction Ø2 equals 62.95° ≈ 63°

Explanation:

The relationship between the angles of incidence and

refraction , with respect to light or other waves passing through two different substances or media, such as glass, water or air is given by Snell's Law.

Snell's Law states that the when light travels from one medium to another, it generally refracts.

It is given by the mathematical expression;

[SinØ1°/SinØ2°] = [n2/n1]

Cross multiplying, we have;

n1 × SinØ1° = n2 × SinØ2°

where, n is the indices of refraction of each substance

Ø is the angle between the ray and the line normal to the surface.

Given the following values;

n1 = 1.36 n2 = 1.31 Ø1 = 30° Ø = ?

n1 × SinØ1° = n1 × SinØ2°

SinØ2° = [n1 × SinØ1°]/n2

substituting the values respectively;

SinØ2° = [1.36 × Sin30°]/1.31

SinØ2° = [1.36 × 0.5]/1.31

SinØ2° = 0.68 × 1.31

SinØ2° = 0.8906

Ø2° = Sin–¹(0.8906)

Ø2° = 62.95° ≈ 63°

4 0
3 years ago
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