Answer:
≈ 192.7 m/s
Step-by-step explanation:
Let's start by finding the velocity of the block just before it hits the ground. We can use the conservation of energy:
- The initial potential energy of the block is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the table.
- The final kinetic energy of the block is (1/2)mv^2, where v is the velocity of the block just before it hits the ground.
Conservation of energy tells us that these two energies are equal:
mgh = (1/2)mv^2
Solving for v, we get:
v = sqrt(2gh)
Plugging in the given values, we get:
v = sqrt(2 * 9.8 m/s^2 * 1.9 m) = 6.06 m/s
Now, let's use conservation of momentum to find the initial speed of the bullet. We know that the total momentum of the system (bullet + block) is conserved before and after the collision. Before the collision, the momentum is:
p = mb * vb
where mb is the mass of the block and vb is its initial velocity, which is 0 since it is at rest.
After the collision, the bullet and block move together with a common velocity v. The total momentum is:
p = (mb + m) * v
where m is the mass of the bullet. Since momentum is conserved:
mb * vb = (mb + m) * v
Solving for vb, we get:
vb = (mb + m) * v / mb
Plugging in the given values, we get:
vb = (345 g + 7 g) / 7 g * 6.06 m/s = 192.7 m/s
Therefore, the initial speed of the bullet was approximately 192.7 m/s.