Question

A 12-g bullet moving horizontally strikes and remains in a 3.0-kg block initially at rest on the edge of a table. The block, which is initially 80 cm above the floor, strikes the floor a horizontal distance of 120 cm from its initial position. What was the initial speed of the bullet?

Answer 1

The initial speed of the bullet is found by using the conservation of momentum principle and the measured distance the block travels after collision. After calculating the fall time of the block using kinematics, the combined velocity post-collision is found and used to compute the bullet's initial speed.

Step-by-step explanation:

To calculate the initial speed of the bullet, we can use the principle of conservation of momentum, since no external forces are acting on the system of bullet and block in the horizontal direction. Before the collision, only the bullet has horizontal momentum. Using the given masses and the distance traveled by the block after the collision, we can find the horizontal speed of the block after collision, from which we can then calculate the bullet's initial speed.

Firstly, to determine the speed of the block right after impact, we use the time it takes to hit the ground to determine its horizontal velocity. Since the block falls 80 cm (0.8 m) under gravity and without initial vertical velocity, we can use the formula (1/2)gt^2 = h to find the time t

1. Calculate the time: t = sqrt(2h/g), where g=9.8 m/s^2, and h=0.8 m.
2. Calculate the speed of the block after the collision: Vb = d/t, where d is the horizontal distance traveled (1.2 m).
3. Apply conservation of momentum: m_bullet * v_bullet_initial = (m_bullet + m_block) * Vb
4. Solve for v_bullet_initial