Respuesta :
Answer:
1) v = 1m/s
2) K.E = 3 kJ
3) [tex]K.E_{2} < K.E_{} < K.E_{1}[/tex]
Explanation:
From the law of conservation of momentum, p = mv
1) For the given question, [tex]p = p_{1} + p_{2}[/tex] ------- eqn 1
substitute mv for p in eqn 1,
[tex]p_{1} = m_{1} v_{1}[/tex]
and [tex]p_{2} = m_{2} v_{2}[/tex]
To solve for p, which is the final momentum after collision
[tex]p = m_{1} v_{1} + m_{2} v_{2}[/tex] ------ eqn 2
Parameters
[tex]m_{1}[/tex] = 1000 kg
[tex]m_{2}[/tex] = 5000 kg
[tex]v_{1}[/tex] = 6 m/s
[tex]v_{2}[/tex] = 0 (static object)
substitute for all m and v in eqn 2
p = [tex](1000 X 6) + (5000 X 0)[/tex]
p = 6000 + 0
p = 6000 kgm/s
Recall p = mv
where p = 6000kgm/s and m = [tex](m_{1} + m_{2})[/tex]
v = ?
6000 kgm/s = (1000 + 5000)v
6000 = 6000v
v = 1m/s
2) K.E = [tex]\frac{1}{2} mv^{2}[/tex]
Therefore, K.E after collision
K.E = [tex]\frac{1}{2} X 6000 X 1^{2}[/tex]
K.E = 0.5 X 6000 X 1 = 3000 J
K.E = 3 kJ
3) Comparing K.Es of the carts, first we write out the K.E of each cart and their combined K.E
(Please ignore the Armstrong in the following equation, I don't know how it appeared from the equation tool)
[tex]K.E_{1} = \frac{1}{2} X 1000 X 6^{2}[/tex] = 18 kJ
[tex]K.E_{2} = \frac{1}{2} X 5000 X 0^{2}[/tex] = 0
Combined K.E = 3 kJ
From the above K.Es,
i) The K.E of the combined cart after collision is lesser than the K.E of the first cart of 1000 kg.
3 kJ < 18 kJ
ii) The K.E of the combined cart is greater than the K.E of the second cart of 5000 kg.
0 kJ < 3 kJ
Conclusion;
[tex]K.E_{2} < K.E_{} < K.E_{1}[/tex]
i.e 0 kJ < 3 kJ < 18 kJ
The speed of the combined carts after the collision is 1 m/sec and the kinetic energy of the combined carts after the collision is 3 KJ and the compairision of kinetic energy is given below.
Given :
Mass of empty cart = 1000 Kg
Empty cart speed = 6 m/sec
Stationary loaded cart having a total mass of 5000 Kg.
a) From the law of conservation of momentum,
[tex]\rm p = p_1+p_2 +...[/tex]
[tex]\rm p= m_1v_1 +m_2v_2[/tex] --- (1)
Where,
[tex]\rm m_1 = 1000\; Kg[/tex]
[tex]\rm m_2 = 5000\;Kg[/tex]
[tex]\rm v_1 = 6\;m/sec[/tex]
[tex]\rm v_2= 0\;m/sec[/tex]
Now put the above values in equation (1) we get,
[tex]\rm p = (1000\times 6)+(5000 \times 0)[/tex]
p = 6000 Kg.m/sec
Now,
p = mv
6000 = (1000 + 5000) v
v = 1 m/sec
b) We know that kinetic energy is,
[tex]\rm KE = \dfrac{1}{2}mv^2[/tex]
[tex]\rm KE = \dfrac{1}{2}\times (1000+5000)\times 1^2[/tex]
KE = 3000 J = 3 KJ
c) Initial KE of empty cart,
[tex]\rm KE = \dfrac{1}{2}\times 1000 \times 6^2[/tex]
KE = 18000 J = 18 KJ
Combined KE = 3 KJ
From the above kinetic energy we conclude that:
KE of the combined cart after collision is lesser than the KE of the first cart of 1000 kg.
3 KJ < 18 KJ
KE of the combined cart is greater than the KE of the second cart of 5000 kg.
0 KJ < 3 KJ
For more information, refer the link given below
https://brainly.com/question/18461965
