Question

A particle of mass m moving in the 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 , then percentage loss in the energy during the collision is close to

(a)44% (b)50% (c) 56% (d) 62% ​

Answer 1

Answer:

44%

Explanation:

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

Answer 2

Answer:

Percentage loss in the energy during the collision is 56%

Explanation:

central idea:- Conservation of linear momentum can be applied but energy is not conserved

consider the movement of two particles as shown above

conserving linear momentum in x- direction

• (Pᵢ)x = (Pf)x

• or, 2mv = (2m+m)vₓ

• Vₓ = 2/3v

Conservation of linear momentum in y-direction

(pᵢ)y = (pf)y

(2mv) = (2m+m)vy

Vy = 2/3v

initial Kinetic energy of the two particles system is

• Eᵢ = 1/2m(2v)²+ 1/2(2m)v²
• 1/2×4mv² + 1/2× 2mv²
• 2mv² + mv² = 3mv²

Final energy of the combined two particles system is

• Ef = 1/2(3m)(vₓ² + vy²)

• = 1/2 (3m)[4v²/9+ 4v²/9]

• = 3m/2 [ 8v² /9] = 4mv²/3

lods in the energy ΔE = Eᵢ - Ef

mv² [3 - 4/3] = 5/3 mv²

percentage loss in the energy during the collision

ΔE/E× 100 = 5/3mv² /3 mv² = 5/9×100 ≈ 56% Answer