Answer:
The height is [tex]h =0.0269 \ m[/tex]
The kinetic energy during collision is not conserved
The Mechanical energy during the collision is not conserved
The mechanical energy after the collision is not conserved
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 0.47\ kg[/tex]
The mass of the wad of putty is [tex]m_p = 0.070 \ kg[/tex]
The speed o the wad of putty is [tex]v_p = 5.60 \ m/s[/tex]
The law of momentum conservation can be mathematically represented as
[tex]p_i = p_f[/tex]
Where [tex]p_i[/tex] is the initial momentum which is mathematically represented as
[tex]p_i =m_p * v_p[/tex]
While [tex]p_f[/tex] is the initial momentum which is mathematically represented as
[tex]p_f = (m_b + m_p)v_f[/tex]
Where [tex]v_f[/tex] s the final velocity
So
[tex]m_p v_p = (m_p + m_b) * v_f[/tex]
Making [tex]v_f[/tex] the subject
[tex]v_f = \frac{m_p v_p}{m_b +m_p}[/tex]
substituting values
[tex]v_f = \frac{(0.070)*(5.60)}{0.47 + 0.070}[/tex]
[tex]v_f = 0.726 \ m/s[/tex]
According to the law of energy conservation
[tex]KE = PE[/tex]
Where KE is the kinetic energy of the system which is mathematically represented as
[tex]KE = \frac{1}{2} (m_p + m_b)v_f^2[/tex]
And PE is the potential energy of the system which is mathematically represented as
[tex]PE = (m_p +m_b) gh[/tex]
So
[tex]\frac{1}{2} (m_p + m_b)v_f^2 = (m_p +m_b) gh[/tex]
Making h the subject of the formula
[tex]h = \frac{v_f^2}{2g}[/tex]
substituting values
[tex]h = \frac{(0.726 )^2 }{2 * 9.8}[/tex]
[tex]h =0.0269 \ m[/tex]
Now the kinetic energy is conserved during collision because the system change it height during which implies some of the kinetic energy was converted to potential energy during collision
The the mechanical energy of the system during the collision is conserved because this energy consists of the kinetic and the potential energy.
Now after the collision the mechanical energy is not conserved because the external force like air resistance has reduced the mechanical energy of that system
