Question

Two 2.0kg bodies A and B collide The velocities before the collision are U1=15i+30j and U2=-10j+5.0j After the collision V1=-5.0j+20j All speeds are given in meters per seconds find the final velocity and how much kinetic energy is gained or lost in the collision

Answer 1

Use the law of conservation of momentum. Since the momentum is a linear measure, we can treat each of the dimension separately:

i-direction:

`m_1u_(1i)+m_2u_(2i)=m_1v_(1i)+m_2v_(2i)\\v_(2i) = (m_1u_(1i)+m_2u_(2i)-m_1v_(1i))/(m_2)=((2\cdot 15-2\cdot10+2\cdot5)kg(m)/(s))/(2kg)=10(m)/(s)`

j-direction:

`m_1u_(1j)+m_2u_(2j)=m_1v_(1j)+m_2v_(2j)\\v_(2j) = (m_1u_(1j)+m_2u_(2j)-m_1v_(1j))/(m_2)=((2\cdot 30+2\cdot5-2\cdot20)kg(m)/(s))/(2kg)=15(m)/(s)`

Answer: Final velocity is: (10i + 15j) m/s

Change in the kinetic energy:

`\Delta E_k = E_(ku)-E_(kv) = (1)/(2)m(u_1^2+u_2^2-v_1^2-v_2^2)=\\=(1)/(2)2kg(1125+125-425-325)(m^2)/(s^2)=500J`

Answer: The system lost 500J worth of kinetic energy in the collision