Question

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are

Answer 1

Answer:

Explanation:

Given

mass of body A $m_a=2 kg$

mass of body $m_b=2 kg$

Velocity before Collision is $u_a=15\hat{i}+30\hat{j}$

$u_b=-10\hat{i}+5\hat{j} m/s$

after collision $v_a=-5\hat{i}+20\hat{j} m/s$

let $v_b_x$ and $v_b_y$ velocity of B after collision in x and y direction

conserving momentum in x direction

$m_a\times 15+m_b\times (-10)=m_a\times (-5)+m_b\times (v_b_x)$

as $m_a=m_b$ thus

$15-10=-5+v_b_x$

$v_b_x=10 m/s$

Conserving momentum in Y direction

$m_a\times 30+m_b\times 5=m_a\times 20+m_b\times (v_b_y)$

$30+5=20+v_b_y$

$v_b_y=15 m/s$

thus velocity of B after collision is

$v_b=10\hat{i}+15\hat{j}$

(b)Change in total Kinetic Energy

Initial Kinetic Energy of A  And B

$K.E._a=\frac{1}{2}\times 2(\sqrt{15^2+30^2})^2=1125 J$

$K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+5^2})^2=125 J$

Total initial Kinetic Energy =1250 J

Final Kinetic Energy of A  And B

$K.E._a=\frac{1}{2}\times 2(\sqrt{5^2+20^2})^2=425 J$

$K.E._b=\frac{1}{2}\times 2(\sqrt{10^2+15^2})^2=325 J$

Final Kinetic Energy $=425+325=750 J$

Change $\Delta K.E.=1250-750=500 J$