Answer:
a
The total initial momentum of the two-block system is  [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
b
The magnitude of the final velocity of the two-block system [tex]v_f = 1.9014 \ m/s[/tex]
c
 the change ΔK=Kfinal−Kinitial in the two-block system's kinetic energy due to the collision is Â
  [tex]\Delta KE =- 847.08 \ J[/tex]
Explanation:
From the question we are told that
  The mass of first  block  is [tex]m_1 = 2.50 \ kg[/tex]
   The initial velocity of first  block is [tex]u_1 = 27.0 \ m/s[/tex]
     The mass of second block is  [tex]m_2 = 33.0\ kg[/tex]
     initial velocity of second block is  [tex]u_2 = 0 \ m/s[/tex]
    Â
The magnitude of the of the total initial momentum of the two-block system is mathematically repented as
    [tex]p_i = (m_1 * u_1 ) + (m_2 * u_2)[/tex]
substituting values
    [tex]p_i = (2.50* 27 ) + (33 * 0)[/tex]
    [tex]p_t = 67.5 \ kg \cdot m/s^2[/tex]
According to the law of linear momentum conservation
    [tex]p_i = p_f[/tex]
Where  [tex]p_f[/tex] is the total final momentum of the system which is mathematically represented as
    [tex]p_f = (m_+m_2) * v_f[/tex]
Where [tex]v_f[/tex] is the final velocity of the system
   [tex]p_i = (m_1 +m_2 ) v_f[/tex]
substituting values
    [tex]67.5 = (2.50+33 ) v_f[/tex]
    [tex]v_f = 1.9014 \ m/s[/tex]
The change in kinetic energy is mathematically represented as
   [tex]\Delta KE = KE_f -KE_i[/tex]
Where [tex]KE_f[/tex] is the final kinetic energy of the two-body system  which is mathematically represented as
    [tex]KE_f = \frac{1}{2} (m_1 +m_2) * v_f^2[/tex]
substituting values
    [tex]KE_f = \frac{1}{2} (2.50 +33) * (1.9014)^2[/tex]
    [tex]KE_f =64.17 J[/tex]
While [tex]KE_i[/tex] is the initial kinetic energy of the two-body system
   [tex]KE_i = \frac{1}{2} * m_1 * u_1^2[/tex]
substituting values
    [tex]KE_i = \frac{1}{2} * 2.5 * 27^2[/tex]
    [tex]KE_i = 911.25 \ J[/tex]
So
  [tex]\Delta KE = 64.17 -911.25[/tex]
 [tex]\Delta KE =- 847.08 \ J[/tex]
