Question

Two 6.7 kg bodies, A and B, collide. The velocities before the collision are and . After the collision, . 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 1

Question:

Two 6.7 kg bodies, A and B, collide. The velocities before the collision are $40i+37j$ and $19 i +3.7 j$ . After the collision $9.8 i + 5.1 j$, $50 i+35.6 j$ . 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,

$u_A=40i+37j=\sqrt{40^2+37^2}=54.5 m/s$

$u_B=19 i +3.7 j=\sqrt{19^2+3.7^2}=19.4m/s$

$v_A=9.8 i + 5.1 j=\sqrt{9.8^2+5.1^2}=11m/s$

$v_B=50 i+35.6 j=\sqrt{50^2+35.6^2}=61.4m/s$

Change in kinetic energy = final kinetic energy - initial kinetic energy

$\Delta K.E.=\frac{1}{2} m(v_A+v_B)^2 -\frac{1}{2} m(u_A+u_B)^2$

$\Delta K.E.=\frac{1}{2} 6.7(11+61.4)^2 -\frac{1}{2} 6.7(54.5+19.4)^2\\\Delta K.E. =17559.89-18295.05\\\Delta K.E. = -735.16 J$