A block of wood (who is suspended by a rope) with a mass of 2Kg moves 10cm from its original position after being struck by a 12g bullet, what's the velocity of the bullet?
How long is the rope or do you mean it swung 10 cm vertically up?
Use the law of conservation of momentum to get the bullet velocity. You will need the velocity of the block after impact, before it swings upward.
You need to be more specific about whether the 10 cm of motion is vertical or horizontal. If it moves H = 10 cm up, then you can use conservation of energy to calculate the velocity Vf of the wood block right after impact.
g H = (1/2) Vf^2
If it moves 10 cm horizontally, you need to know the length of the rope to calculate how far up it swings.
In any case,
m v = (M + m) Vf
where v = bullet velocity
m = bullet mass
M = block mass
Vf = block's velocity after impact
You also need to assume that the bullet stays in the block of wood.
Now once you figure out what direction that 10 cm is in, figure out how high the block rises upwards. I will call it h because I do not have your picture.
then the Potential energy of the target at top of swing = m g h = (2.012)(9.8)h
that is the kinetic energy after the impact
(1/2) m v^2 = (1/2)(2.012) v^2 = (2.012)(9.8)h
solve for v of the block and mass after collision
then
momentum before = momentum after
Mbullet Vbullet = 2.012 v
.012 Vbullet = 2.012 v
you know v so solve for Vbullet
What is the maximum height reached by the ball? If a baseball is projected upward from ground level with an initial velocity of 64 feet per second, then its height is a function of time, given by s= -16t^2 + 64t
To find the velocity of the bullet, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
The total momentum (p) is the sum of the momenta of the block and the bullet. The momentum of an object is given by the product of its mass (m) and its velocity (v).
Let's consider the block as object 1 and the bullet as object 2.
Mass of the block (m1) = 2 kg
Mass of the bullet (m2) = 12 g = 0.012 kg (converting grams to kilograms)
Velocity of the block before the collision (v1) = 0 (since it is suspended and not moving initially)
Velocity of the bullet before the collision (v2) = unknown (let's call it v)
Since the block is initially at rest, the total momentum before the collision is zero:
p(before) = m1 * v1 + m2 * v2 = 0
After the collision:
Let's assume the final velocity of the block (v1f) is negligible (since it moves only by a small distance).
The total momentum after the collision is the momentum of the bullet:
p(after) = m2 * v2
Since the total momentum is conserved, we have: p(before) = p(after)
Therefore, we can write the equation: m1 * v1 + m2 * v2 = m2 * v2
Substituting the known values: 2 kg * 0 + 0.012 kg * v = 0.012 kg * v
Simplifying the equation, we find:
0 = 0.012 kg * v
To solve for the velocity of the bullet (v), we know that any value multiplied by zero is zero. Thus, the velocity of the bullet (v) could be any value.
Therefore, based on the given information, we cannot determine the velocity of the bullet.