A body of mass 4.2kg moving with velocity 10ms-1 due east, hits a stationary body of mass 2.8kg. If they stick together after collision and move with veloci...

Answer Details

In this problem, we can use the principle of conservation of momentum, which states that the total momentum of a system remains constant if no external forces act on it. The initial momentum of the system before the collision is given by: p1 = m1 * v1 where m1 is the mass of the first body (4.2 kg) and v1 is its velocity (10 m/s). p1 = 4.2 kg * 10 m/s p1 = 42 kg m/s The second body is stationary, so its initial momentum is zero: p2 = 0 kg m/s The total initial momentum of the system is therefore: p_initial = p1 + p2 p_initial = 42 kg m/s After the collision, the two bodies stick together and move with a common velocity V. The total mass of the system is the sum of the masses of the two bodies: m_total = m1 + m2 m_total = 4.2 kg + 2.8 kg m_total = 7 kg The final momentum of the system is: p_final = m_total * V According to the principle of conservation of momentum, the total initial momentum of the system is equal to the total final momentum of the system: p_initial = p_final Substituting the values we have found, we get: 42 kg m/s = 7 kg * V Solving for V, we get: V = 42 kg m/s / 7 kg V = 6 m/s Therefore, the value of V after the collision is 6 m/s due east. Thus, the correct answer is "6ms^-1".