Respuesta :
Question:
Two 6.7 kg bodies, A and B, collide. The velocities before the collision are [tex]40i+37j[/tex] and [tex]19 i +3.7 j[/tex] . After the collision [tex]9.8 i + 5.1 j[/tex], [tex]50 i+35.6 j[/tex] . What are (a) the x-component and (b) the y-component of the final velocity of B? (c) What is the change in the total kinetic energy (including sign)?
Answer:
(a) 50 m/s
(b) 35.6 m/s
(c) -735.16 J
Explanation:
Given,
[tex]u_A=40i+37j=\sqrt{40^2+37^2}=54.5 m/s[/tex]
[tex]u_B=19 i +3.7 j=\sqrt{19^2+3.7^2}=19.4m/s[/tex]
[tex]v_A=9.8 i + 5.1 j=\sqrt{9.8^2+5.1^2}=11m/s[/tex]
[tex]v_B=50 i+35.6 j=\sqrt{50^2+35.6^2}=61.4m/s[/tex]
Change in kinetic energy = final kinetic energy - initial kinetic energy
[tex]\Delta K.E.=\frac{1}{2} m(v_A+v_B)^2 -\frac{1}{2} m(u_A+u_B)^2\\[/tex]
[tex]\Delta K.E.=\frac{1}{2} 6.7(11+61.4)^2 -\frac{1}{2} 6.7(54.5+19.4)^2\\\Delta K.E. =17559.89-18295.05\\\Delta K.E. = -735.16 J[/tex]
