Question

A particle of mass m moving in x-direction with speed 2v is hit by another particle of mass 2m moving in the y-direction with speed v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

(A) 44%
(B) 50%
(C) 56%
(D) 62%

Answer 1

In a perfectly inelastic collision between two objects, they stick together and move as one after the collision. The percentage loss in energy during the collision is approximately 44%. The correct answer is (A) 44%

Step-by-step explanation:

In a perfectly inelastic collision, two objects stick together and move as one after the collision. The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.

Since the initial momentum in the x-direction is 2mv + 0 and the final momentum after the collision is (m + 2m)V, we can set up the equation 2mv = (m + 2m)V to solve for V.

Solving for V, we get V = 2v/3.

To find the initial kinetic energy, we use the formula K = (1/2)mv², where m is the mass of the object and v is its velocity.

The initial kinetic energy is K = (1/2)m(2v)² = 2mv².

The final kinetic energy after the collision is K' = (1/2)(3m)(2v/3)² = (2/9)mv².

The percentage loss in energy is given by (K - K')/K * 100%.

Substituting the values of K and K', we have (2mv² - (2/9)mv²)/(2mv²) * 100%, which simplifies to (4/9) * 100%, or approximately 44%, so the answer is (A) 44%.