Question

A block of mass 0.50 kg is moving with a speed of 2 m/s on a smooth surface. it strikes another mass of 1 kg and then they move together as a single body. The energy loss during the collision is?

Answer 1

Answer:

The energy loss during the collision is 0.6667 joules

Explanation:

To answer this question, we need to get the speed with which the blocks move.

We will apply the concept of momentum.

Momentum = Mass × Velocity

Initial momentum = final momentum

Initial momentum = (0.50 × 2) + (0 × 1) = 1kgm/s

Final momentum; Let v be the final speed.

Momentum = v × (1 + 0.5) = 1.5v

1.5v = 1

v = 1/1.5 = 0.6667 m/s

Here we have kinetic energy.

Kinetic energy = 1/2mv²

Initial kinetic energy = 1/2 × 0.50 × 2² = 1 joule

Final kinetic energy = 1/2 × (1.5) × 0.6667² = 0.3333 Joules

The energy loss during the collision is given by:

= 1 - 0.3333 = 0.6667 joules

Answer 2

Explanation:

$\tt energy \: \: lossed \: \large \tt during \: \: collision = ⅔ j$