a cannon of mass m1=8000 kg located on a smooth horizontal platform fires a shell of mass m2=600 kg in horizontal direction with a velocity v2=200m/s.find the velocity of the cannon after it is Shot
Answers
Given :
Mass of cannon = 8000kg
Mass of shell = 600kg
Velocity of shell = 200m/s
To Find :
We have to find velocity of the cannon after shell is shot
Concept :
★ Since no net force acts on the whole system, linear momentum is conserved.
Initial momentum of system will be equal to final momentum of system.
Both cannon and shell are at rest in initial so initial momentum of system will be zero.
Calculation :
➠ m₁v₁ + m₂v₂ = 0
➠ 8000v₁ + (600)(200) = 0
➠ v₁ = -120000/8000
➠ v₁ = -15m/s
Negative sign shows opposite direction.
Cheers!
Answer :
➥ The velocity of the cannon = 15 m/s (Backward)
Given :
➤ Mass of a cannon = 8000 kg
➤ Mass of a shell = 600 kg
➤ Velocity of shell = 200m/s
To Find :
➤ Velocity of the cannon = ?
Required Solution :
In the above Question, we are provided that Mass of cannon is 800 kg, Mass of shell is 600 kg, Initial velocity of cannon is 0 m/s because the cannon are at rest in Initial so initial velocity is 0 m/s and Final velocity of shell is 200 m/s
Now, we have also a Initial velocity,
- Initial velocity of a cannon = 0 m/s
We can find velocity of the cannon by using the linear momentum conservation theorem, There is no external force to the system. Therefore we can use linear momentum conservation theorem which says:
★ m₁v = m₁v₁ + m₂v₂ ★
Here,
- m₁ is the Mass of cannon in kg.
- v is the Initial velocity of the cannon in m/s.
- v₁ is the Final velocity of cannon in m/s.
- m₂ is the Mass of shell in kg.
- v₂ is the Velocity of cannon in m/s.
So let's find velocity (v₁) !
⇛ m₁v = m₁v₁ + m₂v₂
⇛ 8000 × 0 = 8000 × v₁ + 600 × 200
⇛ 0 = 8000 × v₁ + 600 × 200
⇛ 0 = 8000v₁ + 600 × 200
⇛ 0 = 8000v₁ + 120000
⇛ 0 - 8000v₁ = 120000
⇛ -8000v₁ = 120000
⇛ 8000v₁ = -120000
⇛ v₁ = -120000/8000
⇛ v₁ = -120/8
⇛ v₁ = -15 m/s
║Hence, the velocity of of the cannon 15 m/s Backward.║