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 

Options

A. 0.16J
B. 1.00 J
C. 0.67J
D. 0.34 J

Answer

C

Explanation

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`.