A block of mass 0.50 kg is moving with a speed of 2.00 m s^(-1) on a smooth surface. It strikes another mass of 1.00 kg which is at rest and then they move together as a single body. The energy which is lost during the collision is

A block of mass 0.50 kg is moving with a speed of 2.00 m s^(-1) on a s

1.	A block of mass 0.50 kg is moving with a speed of 2.00 m s^(-1) on a smooth surface. It strikes another mass of 1.00 kg which is at rest and then they move together as a single body. The energy which is lost during the collision is
Answer» 0.16J 1.00 J 0.67J 0.34 J Solution :Here, `m_1 = 0.50 kg, u_1 = 2.00 MS^(-1)` `m_2 = 1.00 kg, u_2 = 0` If v is the velocity of the combination after collision , then according to the PRINCIPLE of conservation of linear momentum `m_1 u_1 + m_2u_2 = (m_1 + m_2) v` or `v = (m_1 u_1)/(m_1 + m_2) = (0.50 XX 2.00)/(0.50 + 1.00) = (1.00)/(1.50) = 2/3 ms^(-1) ( :. u_2= 0)` Energy loss = Initial energy - Final energy `=1/2 m_1 u_1^2 - 1/2 (m_1 + m_2)v^2` `= 1/2 xx 0.50 xx (2.00)^2 - 1/2 xx (0.50 + 1.00) xx (2/3)^2` `1.00 - (1.50)/2 xx 4/9` = 0.67 J`.