Question

A bullet of mass 11.1 g is fired into an initially stationary block and comes to rest in the block. The block, of mass 1.01 kg, is subject to no horizontal external forces during the collision with the bullet. After the collision, the block is observed to move at a speed of 5.30 m/s.(a) Find the initial speed of the bullet.(b) How much kinetic energy is lost?

Answer 1

Answer:

a) The initial speed of the bullet is 488 m/s

b) The loss of kinetic energy is 1.3 × 10³ J.

Explanation:

Hi there!

To solve this problem we have to use the conservation of momentum:

initial momentum of the bullet + initial momentum of the block =

final momentum of the block-bullet system

The momentum of an object is calculated as follows:

p = m · v

Where:

p = momentum

m = mass of the object.

v = velocity.

Then, in our system:

p₁₁ = initial momentum of the bullet.

p₂₁ = initial momentum of the block.

p₃₂ = final momentum of the block-bullet system.

p₁₁ + p₂₁ =  p₃₂

The initial momentum of the bullet will be:

p₁₁ = m · v

p₁₁ = 0.0111 kg · v

The initial momentum of the block will be:

p₂₁ = 1.01 kg · 0 m/s = 0 kg · m/s

The final momentum of the block-bullet system will be:

p₃₂ = (1.01 kg + 0.0111 kg) · 5.30 m/s

Then, by conservation of the momentum:

initial momentum of the bullet = momentum of the block-bullet system

0.0111 kg · v = (1.01 kg + 0.0111 kg) · 5.30 m/s

v = ((1.01 kg + 0.0111 kg) · 5.30 m/s)/ 0.0111 kg

v = 488 m/s

The initial speed of the bullet is 488 m/s

b) The initial kinetic energy (KE) of the system is the kinetic energy of the bullet because the block is at rest:

KE = 1/2 · m · v²

KE = 1/2 · 0.0111 kg · (488 m/s)²

KE = 1.32 × 10³ J

The final kinetic energy of the system will be the kinetic energy of the block-bullet system:

KE = 1/2 · (1.01 kg + 0.0111 kg) · (5.30 m/s)²

KE = 14.3 J

The loss of kinetic energy will be:

initial kinetic energy - final kinetic energy

1.32 × 10³ J - 14.3 J = 1.3 × 10³ J

The loss of kinetic energy is 1.3 × 10³ J.