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.
