A bullet of mass 8 gram is fired into a block of mass 2,5 kg initially at rest at the edge of a frictionless table of height 1 m. The bullet remains in the block, and after impact the block lands a distance 2 m from the bottom of the table. Determine the initial speed of the bullet.
first figure out how long it took the block to fall to the floor:
h = (1/2) g t^2
t = sqrt (2 h/g)
so it went 2 meters in t seconds
so the horizontal speed was
u = 2/t
NOW you have a conservation of momentum problem
u(2.5 + .008) = .008 v
solve for v
To determine the initial speed of the bullet, we can use the principle of conservation of momentum.
The momentum of an object is defined as the product of its mass and its velocity. According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Let's call the initial speed of the bullet "v" and the velocity of the block and bullet together after the collision "V". Since the block is initially at rest, its initial velocity is 0.
The total momentum before the collision is given by the equation:
(mass of bullet) × (initial velocity of bullet) + (mass of block) × (initial velocity of block) = (mass of bullet + mass of block) × (velocity after collision)
Using the given values:
(0.008 kg) × (initial velocity of bullet) + (2.5 kg) × (0) = (2.508 kg) × V
Since the bullet remains in the block after the collision, the final velocity of the block (V) can be calculated using the equation for conservation of energy:
Potential energy at the top of the table = Kinetic energy at the bottom of the table
(mass of block + mass of bullet) × g × h = 0.5 × (mass of block + mass of bullet) × V^2
Substituting the given values:
(2.508 kg) × 9.8 m/s^2 × 1 m = 0.5 × (2.508 kg) × V^2
Simplifying the equation:
24.5292 = 1.254 × V^2
V^2 = 19.579
Taking the square root of both sides:
V = 4.42 m/s
Now, we can substitute the value of V into the momentum equation to find the initial velocity of the bullet:
(0.008 kg) × (initial velocity of bullet) + (2.5 kg) × (0) = (2.508 kg) × (4.42 m/s)
Simplifying the equation:
(0.008 kg) × (initial velocity of bullet) = (2.508 kg) × (4.42 m/s)
Dividing both sides by 0.008 kg:
initial velocity of bullet = (2.508 kg) × (4.42 m/s) / (0.008 kg)
Calculating the result:
initial velocity of bullet = 1377.15 m/s
Therefore, the initial speed of the bullet is 1377.15 m/s.