A bullet with a mass of 0.02 kg is fired horizontally into a block of wood hanging on a string. The bullet sricks in the wood an
d causes it to swing upward to a height of 0.2 m. If the mass of the wood block is 2 kg, what is the initial speed of the bullet?
1 answer:
Answer:
u= 20.09 m/s
Explanation:
Given that
m = 0.02 kg
M= 2 kg
h= 0.2 m
Lets take initial speed of bullet = u m/s
The final speed of the system will be zero.
From energy conservation
1/2 m u²+ 0 = 0+ (m+M) g h
m u²=2 (m+M) g h
By putting the values
0.02 x u² = 2 (0.02+2) x 10 x 0.2 ( take g=10 m/s²)
u= 20.09 m/s
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Answer:
9.73 x 10⁻¹⁰ m
Explanation:
According to Heisenberg uncertainty principle
Uncertainty in position x uncertainty in momentum ≥ h / 4π
Δ X x Δp ≥ h / 4π
Δp = mΔV
ΔV = Uncertainty in velocity
= 2 x 10⁻⁶ x 3 / 100
= 6 x 10⁻⁸
mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg
m = 9 x 10⁻¹⁹
Δp = mΔV
= 9 x 10⁻¹⁹ x 6 x 10⁻⁸
= 54 x 10⁻²⁷
Δ X x Δp ≥ h / 4π
Δ X x 54 x 10⁻²⁷ ≥ h / 4π
Δ X = h / 4π x 1 / 54 x 10⁻²⁷
=
= 9.73 x 10⁻¹⁰ m