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liberstina [14]
3 years ago
10

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

Physics
1 answer:
Paha777 [63]3 years ago
7 0

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

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Answer:

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