Respuesta :
Answer:
The initial velocity of the bullet is
Vo =281.52 m/s
Explanation:
Given that,
Mass of block, M = 0.8kg
Then, weight of the block is
W = Mg = 0.8 ×9.81= 7.848N
Length of cord use to suspend mass is
l = 1.6m
Mass of bullet fired is
m=12g = 12/1000 =0.012kg
The initial velocity of the bullet is Vo
But the block is at rest before the bullet was fired, then, it's initial velocity is Uo =0m/s
When the bullet hits the block it moves to a height of h=0.8m
Tension in cord when the body rise to that height is T = 4.8N
So we want to find the initial velocity is Vo.
Using conservation of momentum
Momentum before collision is equal to the momentum after collision
Momentum is given as p=mv
The momentum before collision of the bullet and the block is
P(initial) = mVo + MUo
P(initial) = 0.012×Vo + 0.8 × Uo
P(initial) = 0.012Vo
After collision, both the block and the mass are moving together, i.e. the inelastic collision
The final momentum after collision is given as
P(final) = (m+M)V1
P(final) =(0.012+0.8)V1
P(final) = 0.812V1
Now, we need to know V1.
To know the angle the mass suspended after being fired.
Check attachment for better view
Then, using trigonometry
Cosθ=adj/hyp
Cosθ=0.8/1.6
Cosθ=0.5
θ=arccos(0.5)
θ=60°
Applying Newton second law of motion at the point 2 along radial direction
ΣF = ma
T — W2•Cosθ = (m+M)a
Where a is radial acceleration and it is given as a=v²/r. r=l
T — (m+M)g•Cosθ = (m+M)a
4.8—(0.012+0.8)×9.81•Cos60=(0.012+0.8)V2²/l
4.8 — 3.983=0.812V2²/1.6
0.876 = 0.812V2²/1.6
0.81714 = 0.5075V2²
V2² = 0.81714/0.5075
V2²= 1.61
V2 = √1.61
V2 = 1.269 m/s
V2 ≈ 1.27m/s
This is the velocity of the object and the bullet at point 2
But, we need V1
Let analyse point 1
Using conservation of energy
i.e total energy at point 1 to energy at point 2.
K1 + U1 = K2 + U2
Potential energy is given as mgh
Kinetic energy is given as ½mv²
Note that U1 is zero because change in height is zero
Then, we have
K1 = K2 + U2
½(m+M)V1² =½(m+M)V2²+(m+M)•g•h
½(0.012+0.8)V1² = ½(0.012+0.8)1.27² + (0.012+0.8)•9.81•0.8
0.406V1² = 0.6548 + 6.373
0.406V1² = 7.0274
V1² = 7.0274/0.406
V1² = 17.309
V1 = √17.309
V1 = 4.16m/s
Then, applying it to the conservation of momentum
P(initial) = P(final)
0.012Vo = 0.812V1
0.012Vo = 0.812 ×4.16
0.012Vo = 3.378
Vo = 3.378/0.012
Vo = 281.52 m/s
Then, the initial velocity of the bullet is 281.52m/s
The initial velocity of the bullet is, Vo =281.52 m/s using momentum.
Given:
Mass of block, M = 0.8kg
Weight of the block is , W = Mg = 0.8 ×9.81= 7.848N
Length, l = 1.6m
Mass of bullet fired, m=12g = 12/1000 =0.012kg
To find:
The initial velocity of the bullet is Vo=?
The Calculation for initial velocity:
But the block is at rest before the bullet was fired, then, it's initial velocity is U₀ =0m/s
When the bullet hits the block it moves to a height of h=0.8m
Tension in the cord when the body rise to that height is T = 4.8N
Using conservation of momentum
Momentum before collision is equal to the momentum after collision
Momentum is given as p=mv
The momentum before collision of the bullet and the block is
P(initial) = mV₀ + MU₀
P(initial) = 0.012×V₀ + 0.8 × U₀
P(initial) = 0.012V₀
After collision, both the block and the mass are moving together, i.e. the inelastic collision
The final momentum after collision is given as
P(final) = (m+M)V₁
P(final) =(0.012+0.8)V₁
P(final) = 0.812V₁
For initial velocity:
Cosθ=adj/hyp
Cosθ=0.8/1.6
Cosθ=0.5
θ=arccos(0.5)
θ=60°
Applying Newton's second law of motion at the point 2 along radial direction:
ΣF = ma
T — W2•Cosθ = (m+M)a
Where a is radial acceleration and it is given as a=v²/r. r=l
T — (m+M)g•Cosθ = (m+M)a
4.8—(0.012+0.8)×9.81•Cos60=(0.012+0.8)V₂²/l
4.8 — 3.983=0.812V2²/1.6
0.876 = 0.812V2²/1.6
0.81714 = 0.5075V2²
V₂² = 0.81714/0.5075
V₂²= 1.61
V₂ = √1.61
V₂ = 1.269 m/s
V₂ ≈ 1.27m/s
Using conservation of energy
i.e total energy at point 1 to energy at point 2.
K1 + U1 = K2 + U2
Potential energy is given as mgh
Kinetic energy is given as ½mv²
Note that U₁ is zero because the change in height is zero
Then, we have
K₁ = K₂ + U₂
½(m+M)V₁² =½(m+M)V₂²+(m+M)•g•h
½(0.012+0.8)V₁² = ½(0.012+0.8)1.27² + (0.012+0.8)•9.81•0.8
0.406V₁² = 0.6548 + 6.373
0.406V₁² = 7.0274
V₁² = 7.0274/0.406
V₁² = 17.309
V₁ = √17.309
V₁= 4.16m/s
Then, applying it to the conservation of momentum
P(initial) = P(final)
0.012V₀ = 0.812V₁
0.012V₀ = 0.812 ×4.16
0.012V₀ = 3.378
V₀= 3.378/0.012
V₀ = 281.52 m/s
Then, the initial velocity of the bullet is 281.52m/s
Find more information about momentum here:
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