An 8.00 g bullet is fired into a 160 g block that is initially at rest at the edge of a table of 1.00 m height. The bullet remains in the block, and after the impact the block lands d = 2.1 m from the bottom of the table. Determine the initial speed of the bullet.
To determine the initial speed of the bullet, we can make use of the principle of conservation of momentum. In an isolated system like this one (assuming no external forces act on the system), the total momentum before the collision is equal to the total momentum after the collision.
The total momentum before the collision is the sum of the momentum of the bullet and the momentum of the block. The momentum of an object can be calculated by multiplying its mass by its velocity.
Let's assume the initial velocity of the bullet is v, and the final velocity of the bullet-block system is v'.
The momentum of the bullet before the collision is given by:
Momentum_bullet_before = mass_bullet * velocity_bullet
= 8.00 g * v
The momentum of the block before the collision is given by:
Momentum_block_before = mass_block * velocity_block
= 0 * 160 g (since the block is initially at rest)
The momentum of the bullet-block system after the collision is given by:
Momentum_system_after = (mass_bullet + mass_block) * velocity_system_after
= 8.00 g * v'
According to the conservation of momentum, we have:
Momentum_bullet_before + Momentum_block_before = Momentum_system_after
Considering that the block is initially at rest, we can rewrite the equation as:
8.00 g * v = (8.00 g + 160 g) * v'
Now, let's substitute the given values:
0.008 kg * v = (0.008 kg + 0.16 kg) * v'
Simplifying the equation further:
0.008 v = 0.168 v'
Finally, rearranging the equation to solve for v (the initial velocity of the bullet):
v = (0.168 v') / 0.008
To find v', the final velocity of the bullet-block system after the collision, we can use the kinematic equation that relates distance, time, and initial velocity:
d = v' * t + (1/2) * g * t²
Since we know the final displacement of the system from the bottom of the table is 2.1 m, we need to find the time it takes for the block to hit the ground. Since the block falls vertically downward, we can use the equation for displacement with constant acceleration (gravity) to find the time.
d = (1/2) * g * t²
Solving for t:
t = sqrt((2 * d) / g)
Now we can substitute the value of t into the equation for v':
v' = g * t
Given that the height of the table is 1.00 m, the acceleration due to gravity is approximately 9.81 m/s².
Let's substitute the known values into the equation for t:
t = sqrt((2 * 1.00 m) / 9.81 m/s²)
= sqrt(0.2038 s²)
≈ 0.45 s
And substitute the known values into the equation for v':
v' = 9.81 m/s² * 0.45 s
≈ 4.42 m/s
Finally, substitute the values of v' and g (acceleration due to gravity) into the equation for v:
v = (0.168 * 4.42 m/s) / 0.008
≈ 92.4 m/s
Therefore, the initial speed of the bullet is approximately 92.4 m/s.