Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]p = 14033.7 \ kg\cdot m/s[/tex]
b
[tex]v = 1.22 \ m/s[/tex]
c
[tex]m_2 = 9506.7 \ kg[/tex]
Explanation:
From the question we are told that
The mass of the coal car is [tex]m_1 = 4527 \ kg[/tex]
The velocity is [tex]v_i = 3.1 \ m/s[/tex]
The load of coal dropped is [tex]m_2 = 6952 \ kg[/tex]
Generally the momentum of the system before the coal is added is mathematically represented as
[tex]p = m_j * v_i[/tex]
=> [tex]p = 4527 * 3.1[/tex]
=> [tex]p = 14033.7 \ kg\cdot m/s[/tex]
Generally from the law of momentum conservation is mathematically represented as
[tex]m_1 * v_i + m_2 * v_2 = (m_1 + m_2 )v[/tex]
Here [tex]v_2[/tex] is the velocity of the load before it is dropped and the value is 0 m/s
So
=> [tex]v =\frac{m_1 * v_i }{m_1 + m_2}[/tex]
[tex]v = \frac{4527 * 3.1 }{( 4527 + 6952 )}[/tex]
=> [tex]v = 1.22 \ m/s[/tex]
When v = 1m/s and [tex]m_1[/tex] remain the same
Then
=> [tex]1 =\frac{4527 * 3.1 }{4527 + m_2}[/tex]
=> [tex]4527 + m_2 = 14033.7[/tex]
=> [tex]m_2 = 9506.7 \ kg[/tex]
