Question

A 0.500 kg block of wood is dropped from a height of 1.30 meters. A 0.0250 kg bullet is fired upwards at the falling block with a speed of 340. m/s. At what height does the fired bullet momentarily stop the fall of the block? Assume the time for the bullet to come to rest in the block is negligibly short. Show all work and be sure to use the correct number of significant figures in your answer.

Answer 1

Using conservation of momentum and kinematics, it's calculated that the bullet stops the fall of the block at a height of 14.67 meters above the ground.

Step-by-step explanation:

To find the height at which the bullet momentarily stops the fall of the block of wood, we use the principle of conservation of momentum. As the system is isolated and no external forces are considered during the interaction, the momentum of the system before the collision is equal to the momentum after the collision. The following calculations will assume that the upward direction is positive and the downward is negative.

Before the collision:

• Momentum of the block = 0 (as it is dropped and is gravity is acting on it)
• Momentum of the bullet = mass of the bullet × velocity of the bullet = 0.0250 kg × 340 m/s = 8.5 kg·m/s

After the collision, the bullet and the block move together:

• Total mass = mass of the block + mass of the bullet = 0.500 kg + 0.0250 kg = 0.525 kg
• Let V be the final velocity of the combined mass after the collision.

Conservation of momentum:

0 (momentum of block) + 8.5 kg·m/s (momentum of bullet) = 0.525 kg × V

Solving for V:

V = 8.5 kg·m/s / 0.525 kg = 16.19 m/s (to the nearest significant figures)

Now, using the kinematic equation for the motion of the combined mass (bullet + block) from the point of collision to the maximum height reached, where it momentarily stops before falling back down:

Final velocity (Vf) = 0 (at the highest point)

Initial velocity (Vi) = V = 16.19 m/s

Acceleration (a) = -9.8 m/s² (due to gravity, negative as it's opposing the upward motion)

Using the kinematic equation Vf^2 = Vi^2 + 2aΔy

0 = (16.19 m/s)^2 + 2(-9.8 m/s²)Δy

Solving for Δy, the change in height from the point of collision:

Δy = - (16.19 m/s)^2 / (2×-9.8 m/s²) = 13.37 m (to the nearest significant figures)

To find the height at which the bullet stops the block, we need to add this change in height (Δy) to the initial drop height (1.30 m) of the block:

Height = Initial height + Δy = 1.30 m + 13.37 m = 14.67 m

So, the bullet momentarily stops the fall of the block at a height of 14.67 meters above the ground.