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:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.
where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing: