conservation of momentum statement and numerical problem
Answers
Explanation:
Momentum Conservation Principle
The Law of Action-Reaction (Revisited)
Momentum Conservation Principle
Isolated Systems and Collision Analysis
Collision Analysis and Momentum Problems
Using Equations as Guides to Thinking
Momentum Conservation in Explosions
One of the most powerful laws in physics is the law of momentum conservation. The law of momentum conservation can be stated as follows.
For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. That is, the momentum lost by object 1 is equal to the momentum gained by object 2.
The above statement tells us that the total momentum of a collection of objects (a system) is conserved - that is, the total amount of momentum is a constant or unchanging value. This law of momentum conservation will be the focus of the remainder of Lesson 2. To understand the basis of momentum conservation, let's begin with a short logical proof
Question 1: Find the recoil velocity of a gun having mass equal to 5 kg, if a bullet of 25gm acquires the velocity of 500m/s after firing from the gun.
Answer: Here given,
Mass of bullet (m1) = 25 gm = 0.025 kg
Velocity of bullet before firing (u1) = 0
Velocity of bullet after firing (v1) = 500 m/s
Mass of gun (m2) = 5 kg
Velocity of gun before firing, (u2) = 0
Velocity of gun after firing = ?
We know that,
m
1
u
1
+
m
2
u
2
=
m
1
v
1
+
m
2
v
2
⇒
0.025
k
g
×
0
+
5
k
g
×
0
=
0.025
k
g
×
500
m
/
s
+
5
k
g
×
v
2
⇒
0
=
12.5
k
g
m
/
s
+
5
k
g
×
v
2
⇒
5
k
g
×
v
2
=
−
12.5
k
g
m
/
s
⇒
v
2
=
−
12.5
k
g
m
/
s
5
k
g
⇒
v
2
=
−
2.5
m
/
s
Thus, recoil velocity of gun is equal to 2.5 m/s. Here negative (- ve) sign shows that gun moves in the opposite direction of bullet.