.A 5.5kg riffle fires an 11gram bullet with a velocity of 1000m/s.
a. find the velocity of a riffle?
b. if the shooter holds the riffle firmly with his shoulder what will be the velocity assume the mass of shooter is 120kg?
Solution
*Let,
m1= 0.011 kg
v1= 1000 m/s
m2=5.5 kg and
v2= be the recoil velocity
*Equation:
M1V1+M2V2=0
a. (0.011 kg)(1000 m/s) + (5.5 kg)(v2)=0
11 kgm/s + 5.5 kgv2 = 0
5.5 kgv2 = -11 kg
v2= -2 m/s
b. (0.011 kg)(1000 m/s) + (120 kg + 5.5 kg)(v2)=0
11 kgm/s + 125.5 kgv2 = 0
125.5 kgv2 = -11 kgm/s
V2= -0.9 m/s or -9 cm/s
Use conservation of momentum to solve this.
a. To find the velocity of the rifle, we can use the principle of conservation of momentum. The momentum of the bullet before it is fired is equal to the momentum of the bullet plus rifle after it is fired.
Momentum of the bullet before firing = Momentum of the bullet + Momentum of the rifle after firing
Momentum = mass × velocity
The mass of the bullet is given as 11 grams, which is equal to 0.011 kg. The velocity of the bullet is given as 1000 m/s.
Momentum of the bullet before firing = (0.011 kg) × (1000 m/s) = 11 kg·m/s
The mass of the rifle is given as 5.5 kg.
Let the velocity of the rifle be v.
Momentum of the rifle after firing = (5.5 kg) × (v m/s) = 5.5v kg·m/s
According to the conservation of momentum,
Momentum of the bullet before firing = Momentum of the bullet + Momentum of the rifle after firing
11 kg·m/s = 11 kg·m/s + 5.5v kg·m/s
Simplifying the equation,
0 = 5.5v kg·m/s
This implies that the velocity of the rifle is 0 m/s.
Therefore, the velocity of the rifle is 0 m/s.
b. When the shooter holds the rifle firmly with his shoulder, the impulse experienced by the shooter will cause a backward recoil force, which must be balanced by an equal and opposite forward force on the bullet. This means that the momentum transferred to the rifle will cancel out the momentum transferred to the bullet.
Momentum of the shooter = Momentum of the bullet + Momentum of the rifle
Momentum of the shooter = (mass of the shooter) × (velocity of the shooter)
Given that the mass of the shooter is 120 kg and the velocity of the shooter is v (unknown), and the mass and velocity of the bullet are 0.011 kg and 1000 m/s respectively, we can set up the equation as follows:
(120 kg) × (v m/s) = (0.011 kg) × (1000 m/s) + (5.5 kg) × (v m/s)
Simplifying the equation,
120v = 11 + 5.5v
114.5v = 11
v ≈ 0.0961 m/s
Therefore, when the shooter holds the rifle firmly, the velocity of the rifle will be approximately 0.0961 m/s.
To solve this problem, we can use the principle of conservation of momentum. The total initial momentum of the system (rifle and bullet) is equal to the total final momentum of the system after the bullet is fired.
a. Finding the velocity of the rifle:
Let's denote the velocity of the rifle as V (unknown) and the velocity of the bullet as v (1000 m/s). The mass of the bullet (m1) is 11 grams or 0.011 kg, and the mass of the rifle (m2) is 5.5 kg. The mass of the system before firing is 5.511 kg (m1 + m2).
Using the conservation of momentum equation:
Initial momentum = Final momentum
(0 + m2) x 0 = (m1 x v) + (m2 x V)
Substituting the known values:
0 = (0.011 kg x 1000 m/s) + (5.5 kg x V)
Simplifying the equation:
0 = 11 kg·m/s + 5.5 kg x V
Rearranging the equation to solve for V:
V = -11 kg·m/s / 5.5 kg
V = -2 m/s
Therefore, the velocity of the rifle is -2 m/s. The negative sign indicates that the rifle moves in the opposite direction of the bullet.
b. If the shooter holds the rifle firmly with his shoulder:
In this case, we can consider the shooter and rifle as a combined system. The total initial momentum of the system before firing is zero, as there is no net external force acting on the system.
Using the same conservation of momentum equation:
Initial momentum = Final momentum
(0 + m2 + m3) x 0 = (m1 x v) + (m2 x V)
Now, we need to consider the shooter's mass (m3), which is 120 kg. Substituting the known values (m1 = 0.011 kg, m2 = 5.5 kg, m3 = 120 kg, and v = 1000 m/s):
0 = (0.011 kg x 1000 m/s) + (5.5 kg x V) + (120 kg x 0)
Simplifying the equation:
0 = 11 kg·m/s + 5.5 kg x V
Rearranging the equation to solve for V:
V = -11 kg·m/s / 5.5 kg
V = -2 m/s
In this case, the velocity of the rifle is also -2 m/s, irrespective of the mass of the shooter. The shooter's mass does not affect the rifle's velocity. This result is because the shooter and rifle's backward momentum cancel out the forward momentum of the bullet, resulting in a net momentum of zero.