Question

a block of mass 0.50 kg is moving with a speed of 2.00 ms 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

Answer 1

The energy lost during the collision is 0.67 J

Explanation:

Initial kinetic energy of the system

K.E = 1 / 2μ^2 + 1 / 2 M(0)^2

K.E = 1 / 2 × 0.5 × 2 × 2 + 0 = 1 J

For collision , applying conservation of linear momentum

m × u = (m+M) × ν

0.5 × 2 = (0.5+1) × ν

ν = 23 m/s

Final kinetic energy of the system is

K.E(f) = 1 / 2(m+M)ν^2

        = 1 / 2(0.5+1) × 23 × 23 = 13 j

Energy loss during collision = (1−1 / 3) J = 0.67 J

Hence the energy lost during the collision is 0.67 J

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