Answer:
4500 J and 3000 J
Explanation:
According to conservation of momentum
[tex]0 = m_1 V_1 + m_2 V_2[/tex]
Given that m_2 = 1.5 m_1 , so
[tex]V_1 = -1.5 V_2[/tex]
the kinetic energy of each piece is
[tex]K_2= \frac{1}{2} m_2v_2^2[/tex]
[tex]K_1= \frac{1}{2} m_1v_1^2[/tex]
substituting the value of V1 in the above equation
[tex]K_1 = (1/2)( m_2 / 1.5 )( -1.5 V_2)^2 = 1.5 (1/2)m_2 V_2^2 = 1.5 K_2[/tex]
Given that
K_1 + k_2 = 7500 J
1.5 K_2 + K_2 = 7500
K_2 = 7500 / 2.5
= 3000 J
this is the KE of heavier mass
K_1 = 7500 - 3000 = 4500 J
this is the KE of lighter mass
Each piece of the object possess a kinetic energy of 3000 and 4500 Joules respectively.
Given the following data:
To find the amount of kinetic energy possessed by each piece of an object:
Applying the law of conservation of momentum, we have:
[tex]M_AV_A + M_BV_B = 0[/tex]
[tex]M_AV_A = - M_BV_B \\\\M_AV_A = -1.5 M_AV_B\\\\V_A =\frac{ -1.5 M_AV_B}{M_A}\\\\V_A =\frac{ -3 V_B}{2}[/tex]
For the energy:
[tex]Total \;kinetic \;energy = Kinetic \;energy \;A + Kinetic \;energy \;B\\\\Total \;kinetic \;energy = \frac{1}{2}M_AV_A^2 + \frac{1}{2}M_BV_B^2[/tex]
Substituting the values, we have:
[tex]7500 = \frac{1}{2}M_AV_A^2 + \frac{1}{2}(\frac{3M_A}{2}) (\frac{2V_A}{3})^2\\\\7500 = \frac{1}{2}M_AV_A^2 + \frac{3M_A}{4} (\frac{2V_A}{3})^2\\\\7500 = \frac{5}{2} (\frac{1}{2}M_AV_A^2)\\\\7500 = \frac{5}{2} (K.E_A)\\\\5K.E_A = 7500 \times 2\\\\5K.E_A = 15000\\\\K.E_A = \frac{15000}{5}[/tex]
Kinetic energy A = 3000 Joules
For Kinetic energy B:
[tex]Kinetic \;energy\; B = 7500 - 3000[/tex]
Kinetic energy B = 4500 Joules
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