Question

n explosion breaks an object initially at rest into two pieces, one of which has 2.0 times the mass of the other. If 8400 J of kinetic energy were released in the explosion, how much kinetic energy did the heavier piece acquire

Answer 1

The heavier piece acquired 2800J of energy.

Explanation:

Let the velocity of the first piece be v1 and that of the second piece be v1 and also let the masses be m1 and m2 respectively.

m2 = 2m1

From the principles of conservation of momentum

m1v1 = m2v2

m1v1 = 2m1v2

Dividing through by m1

That is v1 = 2v2

v2 = 1/2v1

Total kinetic energy = 8400J

1/2m1v1² + 1/2m2v2² = 8400

1/2m1v1² + 1/2(2m1) × (1/2v1)² = 8400

1/2m1v1² + 1/4m1v1² = 8400

Let 1/2m1v1 = K1

K1 + 1/2K1 = 8400

3/2K1 = 8400

K1 = 8400× 2/3 = 5600J

K2 = 8400 – 5600 = 2800J

Answer 2

The heavier piece acquired 2800 J kinetic energy

Step-by-step explanation:

From the principle of conservation of linear momentum:

0 = M₁v₁ - M₂v₂

M₁v₁ = M₂v₂

let the second piece be the heavier mass, then

M₁v₁ = (2M₁)v₂

v₁ = 2v₂ and v₂ = ¹/₂ v₁

From the principle of conservation of kinetic energy:

¹/₂ K.E₁ + ¹/₂ K.E₂ = 8400 J

¹/₂ M₁(v₁)² + ¹/₂ (2M₁)(¹/₂v₁)² = 8400

¹/₂ M₁(v₁)² + ¹/₄M₁(v₁)² = 8400

K.E₁ + ¹/₂K.E₁ = 8400

Now, we determine K.E₁ and note that K.E₂ = ¹/₂K.E₁

1.5 K.E₁ = 8400

K.E₁ = 8400/1.5

K.E₁ = 5600 J

K.E₂ = ¹/₂K.E₁ = 0.5*5600 J = 2800 J

Therefore, the heavier piece acquired 2800 J kinetic energy