Question

A bomb of mass 30 kg at rest explodes into two pieces of 18 kg and 12 kg developed city of the kinetic energy of the other mass will be

Answer 1

Answer:

486J

Explanation:

Mass of bomb = 30kg (Given)

Explosion into to pieces = 12 and 18 kg (Given)

Initial momentum of the bomb = 0 

After exploding, let the speed of mass 18 kg is 6 m/s and the speed of mass 12 be U

Thus, the momentum after collision = momentum 1 + momentum 2

P = 18 × 6 + 12 × U

P = 108 + 12 U

According to the conservation of momentum -

momentum before collision = momentum after collision

o = 108 + 12 U

-108 = 12U

U = -108/12

U = -9

KE = 1/2 mv²

KE = 1/2 x 12 x -9²

KE = 486J

Thus, the kinetic energy of the other mass will be 486J.

Answer 2

Question:

A bomb of mass 30 kg at rest explodes into 2 pieces of masses 18 kg and 12 kg the velocity of 18 kg mass is 6 m/s the KE of the other mass will be ?

Answer:

486 J

Explanation:

The linear momentum of the exploding part will remain conserved.

Applying Conservation of linear momentum,

m₁u₁ = m₂u₂

here,

m₁ = 18 kg

m₂ = 12 kg

u₁ = 6 m/s

u₂ = ?

u₂ = 18×6÷12 = 9 m/s

u₂ = 9 m/s

Now, The kinetic energy of 12 kg mass

K.E₂ = m₂u₂²/2 = 12×(9)²/2 = 6×81 = 486 J