Respuesta :
(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
[tex]p_i = M v_i =(20.0 kg)(+200.0 m/s)=4000 kg m/s[/tex]
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 [tex]v_2 = -500 m/s[/tex] and its mass is [tex]m_2 = 4.00 kg[/tex]. The third part has unknown velocity [tex]v_x[/tex] on the x-direction, and its mass is the mass of the initial body minus the mass of the other two parts:
[tex]m_3 = 20.0 kg -10.0 kg-4.00 kg = 6.00 kg[/tex]
So, the final momentum on the x-direction is
[tex]p_f = m_2 v_2 + m_3 v_3 [/tex]
Since the momentum on the x-axis must be conserved, we can write
[tex]p_i = p_f[/tex]
[tex]M v_i = m_2 v_2 + m_3 v_3[/tex]
from which we find
[tex]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 [/tex]
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
[tex]p_i = 0[/tex]
Then the object breaks into three parts. The first part has speed [tex]v_1 = +100 m/s[/tex] 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 [tex]v_3[/tex]. So the final momentum along the y-axis is
[tex]p_f = m_1 v_1 + m_3 v_3[/tex]
For the conservation of momentum,
[tex]p_i = p_f[/tex]
[tex]0=m_1 v_1 + m_3 v_3[/tex]
From which we find
[tex]v_3 = - \frac{m_1 v_1}{m_3}= -\frac{(10 kg)(100 m/s)}{6 kg} =-166.7 m/s [/tex]
Therefore, the velocity of the third part in the y-direction is [tex]v_3 = -166.7 m/s[/tex] (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
[tex]K_i = \frac{1}{2} Mv_i^2 = \frac{1}{2}(20.0 kg)(200 m/s)^2=4 \cdot 10^5 J [/tex]
After the explosion, the object is in three parts. The first part has mass [tex]m_1 = 10.0kg[/tex] and speed [tex]v_1 = 100 m/s[/tex]. The second part has mass [tex]m_2 = 4.0 kg[/tex] and speed [tex]v_2 = 500 m/s[/tex]. The third part has mass [tex]m_3 = 6.0 kg[/tex] and its velocity is the resultant of the components on the x- and y-axis:
[tex]v_3 = \sqrt{(1000 m/s)^2+(166.7 m/s)^2}=1013.8 m/s [/tex]
So the total kinetic energy after the explosion is:
[tex]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 = [/tex]
[tex]= \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 = [/tex]
[tex]=3.6 \cdot 10^6 J[/tex]
Therefore, the energy released in the explosion is
[tex]\Delta K = K_f - K_i = 3.6 \cdot 10^6 J - 4 \cdot 10^5 J = 3.2 \cdot 10^6 J[/tex]
- x direction: the body is initially moving on the x-axis in the positive direction, so the initial momentum in the x-direction is
[tex]p_i = M v_i =(20.0 kg)(+200.0 m/s)=4000 kg m/s[/tex]
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 [tex]v_2 = -500 m/s[/tex] and its mass is [tex]m_2 = 4.00 kg[/tex]. The third part has unknown velocity [tex]v_x[/tex] on the x-direction, and its mass is the mass of the initial body minus the mass of the other two parts:
[tex]m_3 = 20.0 kg -10.0 kg-4.00 kg = 6.00 kg[/tex]
So, the final momentum on the x-direction is
[tex]p_f = m_2 v_2 + m_3 v_3 [/tex]
Since the momentum on the x-axis must be conserved, we can write
[tex]p_i = p_f[/tex]
[tex]M v_i = m_2 v_2 + m_3 v_3[/tex]
from which we find
[tex]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 [/tex]
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
[tex]p_i = 0[/tex]
Then the object breaks into three parts. The first part has speed [tex]v_1 = +100 m/s[/tex] 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 [tex]v_3[/tex]. So the final momentum along the y-axis is
[tex]p_f = m_1 v_1 + m_3 v_3[/tex]
For the conservation of momentum,
[tex]p_i = p_f[/tex]
[tex]0=m_1 v_1 + m_3 v_3[/tex]
From which we find
[tex]v_3 = - \frac{m_1 v_1}{m_3}= -\frac{(10 kg)(100 m/s)}{6 kg} =-166.7 m/s [/tex]
Therefore, the velocity of the third part in the y-direction is [tex]v_3 = -166.7 m/s[/tex] (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
[tex]K_i = \frac{1}{2} Mv_i^2 = \frac{1}{2}(20.0 kg)(200 m/s)^2=4 \cdot 10^5 J [/tex]
After the explosion, the object is in three parts. The first part has mass [tex]m_1 = 10.0kg[/tex] and speed [tex]v_1 = 100 m/s[/tex]. The second part has mass [tex]m_2 = 4.0 kg[/tex] and speed [tex]v_2 = 500 m/s[/tex]. The third part has mass [tex]m_3 = 6.0 kg[/tex] and its velocity is the resultant of the components on the x- and y-axis:
[tex]v_3 = \sqrt{(1000 m/s)^2+(166.7 m/s)^2}=1013.8 m/s [/tex]
So the total kinetic energy after the explosion is:
[tex]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 = [/tex]
[tex]= \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 = [/tex]
[tex]=3.6 \cdot 10^6 J[/tex]
Therefore, the energy released in the explosion is
[tex]\Delta K = K_f - K_i = 3.6 \cdot 10^6 J - 4 \cdot 10^5 J = 3.2 \cdot 10^6 J[/tex]
Answer:
Part a)
[tex]\vec v = 1000 \hat i - \frac{500}{3}\hat j[/tex]
Part b)
[tex]\Delta E = 3.23 \times 10^6 J[/tex]
Explanation:
(a) When there is no external force on the system then we can use the law of conservation of momentum
[tex]P_i = P_f[/tex]
Initially the object is moving along x-axis in the positive direction, so the initial momentum in the x-direction is
[tex]P_i = 20(200 \hat i) = 4000 \hat i kg m/s[/tex]
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
[tex]P_f = 10(100 \hat j) + 4(500 -\hat i) + 6(\vec v)[/tex]
Since total momentum is conserved so we can say
[tex]4000 \hat i = 1000\hat j -2000 \hat i + 6 \vec v[/tex]
[tex]\vec v = \frac{4000\hat i + 2000\hat i - 1000\hat j}{6}[/tex]
[tex]\vec v = 1000 \hat i - \frac{500}{3}\hat j[/tex]
b) The energy released is the difference of the kinetic energy of the object before and after the explosion.
[tex]\Delta E = KE_f - KE_i[/tex]
Before the explosion, its kinetic energy is
[tex]KE_i = \frac{1}{2}mv_i^2[/tex]
[tex]KE_i = \frac{1}{2}(20)(200^2)[/tex]
[tex]KE_i = 4\times 10^5 J[/tex]
After the explosion, the object is in three parts. So the total kinetic energy after the explosion is:
[tex]KE_f = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2[/tex]
[tex]KE_f = \frac{1}{2}(10)(100^2) + \frac{1}{2}(4)(500^2) + \frac{1}{2}(6)(1000^2 + 166.67^2)[/tex]
[tex]KE_f = 5\times 10^4 + 5 \times 10^5 + 3.1 \times 10^6[/tex]
[tex]KE_f = 3.63 \times 10^6 J[/tex]
Therefore, the energy released in the explosion is
[tex]\Delta E = 3.63 \times 10^6 - 4 \times 10^5[/tex]
[tex]\Delta E = 3.23 \times 10^6 J[/tex]
