A block of mass 0.50 kg is moving with a speed of 2.00 m/s on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Options

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

Answer

C

Explanation

From law of conservation of momentum, we have
`m_1v + m_2v_2 = (m_1 + m_2)v`
Given `m_1 = 0.50 kg, v_1 = 2 ms^(-1)`
`m_2 = 1 kg, v_2 = 0 " [at rest]`
`0.5 xx 2 + 1 xx 0 = 1.5 xx v" [assumed that" 2^(nd)` body is at rest]
`implies v = 2/3 " ":. " "Delta K = K_t - K_i " " =(1.5 xx (2/3)^2)/(2) - (0.5) xx(2^2)/(2) = -2/3 J`
So, energy lost is 0.67 J.