Question

A cube moving on a smooth horizontal surface collides inelastically with a small obstacle which is fixed on the ground. The percentage loss in kinetic energy in this collision is

40%


25%


62.5%


75.0%

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Answer 1

Explanation:

Momentum should be conserved due to the absence of external force on the system (skater + man).

So from conservation of momentum, we have

m 1 v 1 +m 2v 2 =(m 1 +m 2 )Vwhere V is common velocity after the cling.

we get 60×10+0=100V or V=6m/s

Initial Kinetic Energy of the man and skater was:

K 1 = 21 m 1 v 12+0=30×100=3000Joule

Final Kinetic Energy of the man and skater is

K 2 = 21 (m 1 +m 2 )V 2= 21

100×36=1800Joule

Loss in the kinetic energy will be

ΔKE=3000−1800=1200Joule

Answer 2

Answer:

Kinetic Energy loss is 40%.

Explanation:

In the above question there is a loss of 40% kinetic energy.

When a cube moving on a smooth horizontal surface collides inelastically with a small obstacle which is fixed on the ground, the percentage loss in kinetic energy of the cube can be significant.

This type of collision, the cube and the obstacle will stick together after the impact, and the kinetic energy of the cube will be transferred to the obstacle.

he percentage loss in kinetic energy of the cube can vary depending on the specific circumstances of the collision.

For example, if the cube has a high initial kinetic energy and the obstacle is relatively small, the percentage loss in kinetic energy may be as high as 75%. However, if the cube has a lower initial kinetic energy and the obstacle is larger, the percentage loss in kinetic energy may be closer to 25%.

Overall, the percentage loss in kinetic energy in an inelastic collision between a cube and an obstacle will depend on the relative sizes and initial kinetic energies of the two objects involved.

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