Physics, 23.07.2019 08:30 maybellforever120215

A20.0 kg body is moving through space in the positive direction of an x axis with a speed of 200 m/s when, due to an internal explosion, it breaks into three parts. one part, with a mass of 10.0 kg, moves away from the point of explosion with a speed of 100 m/s in the positive y direction. a second part, with a mass of 4.00 kg, moves in the negative x direction with a speed of 500 m/s. (a) in unit-vector notation, what is the velocity of the third part? (b) how much energy is released in the explosion? ignore effects due to the gravitational force.

Answers: 1

Show answers

Answers

(a) this part of the problem can be solved by using the law of conservation of momentum, applying it on both x- and y- directions.

- x direction: the body is initially moving on the x-axis in the positive direction, so the initial momentum in the x-direction is
p_i = M v_i =(20.0 kg)(+200.0 m/s)=4000 kg m/s

p_i = M v_i =(20.0 kg)(+200.0 m/s)=4000 kg m/s

Then the object breaks into three parts. The first part move in the y-direction, so its velocity has no component on the x-axis. The second part moves in the negative x-direction, so its speed is v_2 = -500 m/s

and its mass is m_2 = 4.00 kg

. The third part has unknown velocity v_x

on the x-direction, and its mass is the mass of the initial body minus the mass of the other two parts:
m_3 = 20.0 kg -10.0 kg-4.00 kg = 6.00 kg

So, the final momentum on the x-direction is
p_f = m_2 v_2 + m_3 v_3

Since the momentum on the x-axis must be conserved, we can write
p_i = p_f

from which we find
$v_3 = \frac{Mv_i - m_2 v_2}{m_3}= \frac{(20kg)(200 m/s)-(4 kg)(-500 m/s)}{6 kg} =+1000 m/s$
Therefore, the velocity of the third part on the x-direction is +1000 m/s (in the positive direction)

- y direction: let's do the same for the y-direction. The initial momentum of the object along the y-direction is zero (because the object was moving along the x-axis), so
p_i = 0

Then the object breaks into three parts. The first part has speed v_1 = +100 m/s

in the y-direction. The second part has speed zero in the y-direction, since it moves along the x-direction. The third part has unknown velocity v_3

. So the final momentum along the y-axis is
p_f = m_1 v_1 + m_3 v_3

For the conservation of momentum,
p_i = p_f

From which we find
$v_3 = - \frac{m_1 v_1}{m_3}= -\frac{(10 kg)(100 m/s)}{6 kg} =-166.7 m/s$
Therefore, the velocity of the third part in the y-direction is v_3 = -166.7 m/s

(in the negative y-direction)

- Now we have the two components of the velocity of the third part in the x- and y-direction, so we can write its velocity in a unit-vector notation:
v = 1000 i - 166.7 j m/s

b) We can solve this part of the problem by comparing the kinetic energy of the object before and after the explosion.

Before the explosion, its kinetic energy is
$K_i = \frac{1}{2} Mv_i^2 = \frac{1}{2}(20.0 kg)(200 m/s)^2=4 \cdot 10^5 J$
After the explosion, the object is in three parts. The first part has mass m_1 = 10.0kg

and speed

. The second part has mass m_2 = 4.0 kg

and speed

. The third part has mass m_3 = 6.0 kg

and its velocity is the resultant of the components on the x- and y-axis:
$v_3 = \sqrt{(1000 m/s)^2+(166.7 m/s)^2}=1013.8 m/s$

So the total kinetic energy after the explosion is:
$K_f = K_1 + K_2 + K_3 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 + \frac{1}{2}m_3 v_3^2 =$
$= \frac{1}{2}(10 kg)(100 m/s)^2 + \frac{1}{2}(4 kg)(500 m/s)^2 + \frac{1}{2}(6 kg)(1013.8 m/s)^2 =$
$=3.6 \cdot 10^6 J$

Therefore, the energy released in the explosion is
$\Delta K = K_f - K_i = 3.6 \cdot 10^6 J - 4 \cdot 10^5 J = 3.2 \cdot 10^6 J$

Answer from: leslymejiaquiroz1

Part a)

$\vec v = 1000 \hat i - \frac{500}{3}\hat j$

Part b)

$\Delta E = 3.23 \times 10^6 J$

Explanation:

(a) When there is no external force on the system then we can use the law of conservation of momentum

P_i = P_f

Initially the object is moving along x-axis in the positive direction, so the initial momentum in the x-direction is

$P_i = 20(200 \hat i) = 4000 \hat i kg m/s$

Then the object breaks into three parts.

The first part move in the y-direction.

The second part moves in the negative x-direction

The third part has unknown velocity

So, the final momentum of all three parts of the body is given as

$P_f = 10(100 \hat j) + 4(500 -\hat i) + 6(\vec v)$

Since total momentum is conserved so we can say

$4000 \hat i = 1000\hat j -2000 \hat i + 6 \vec v$

$\vec v = \frac{4000\hat i + 2000\hat i - 1000\hat j}{6}$

$\vec v = 1000 \hat i - \frac{500}{3}\hat j$

b) The energy released is the difference of the kinetic energy of the object before and after the explosion.

$\Delta E = KE_f - KE_i$

Before the explosion, its kinetic energy is

$KE_i = \frac{1}{2}mv_i^2$

$KE_i = \frac{1}{2}(20)(200^2)$

$KE_i = 4\times 10^5 J$

After the explosion, the object is in three parts. So the total kinetic energy after the explosion is:

$KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$KE_f = \frac{1}{2}(10)(100^2) + \frac{1}{2}(4)(500^2) + \frac{1}{2}(6)(1000^2 + 166.67^2)$

$KE_f = 5\times 10^4 + 5 \times 10^5 + 3.1 \times 10^6$

$KE_f = 3.63 \times 10^6 J$

Therefore, the energy released in the explosion is

$\Delta E = 3.63 \times 10^6 - 4 \times 10^5$

$\Delta E = 3.23 \times 10^6 J$

Answer from: Quest

To use anything transparent

Answer from: Quest

No because a circle or pie graph usually shows percentages you'd need a line graph

Another question on Physics

Physics, 22.06.2019 18:00

Which missing item would complete this alpha decay reaction ? > 228/88ra + 4/2he a.224/96 rn b. 232/90 th c. 230/92 u d. 235/92 u

Answers: 2

Answer

Physics, 22.06.2019 23:10

Ablue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.5 m/s 2 . a green car arrives at the position of the stop-light 6.5 s after the light had turned green. if t = 0 when the light turns green, at what time does the green car catch the blue car if the green car maintains the slowest constant speed necessary to catch up to the blue car? answer in units of s.

Answers: 1

Answer

Physics, 23.06.2019 09:00

When the acceleration of a mass on a spring is zero, the velocity is at a (fill in the blank) . when the velocity of a mass on a spring is zero, the acceleration is at a (fill in the blank) .

Answers: 1

Answer

Physics, 23.06.2019 11:00

Research and discuss the aurora borealis in terms of electric charges, magnetic fields, and forces. which times of the year are best for seeing the northern lights, and where are the best places to view them? explain your answers. also identify links to good images. if you’ve ever seen the aurora borealis, describe your experience and note the time and place that you saw it.

Answers: 2

Answer

You know the right answer?

A20.0 kg body is moving through space in the positive direction of an x axis with a speed of 200 m/s...

Questions