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.

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(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

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

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

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