Question

If a 0.75kg bullet made a 3.0kg block swing to change its height 0.5 Meters.

How fast was the bullet traveling?

Answer 1

To find the velocity of the bullet, the potential energy gained by the block when raised 0.5 meters was equated to the kinetic energy of the bullet. The bullet was traveling at approximately 6.26 m/s before it collided with the block.

Step-by-step explanation:

To determine how fast a 0.75kg bullet was traveling before it made a 3.0kg block swing and change its height by 0.5 meters, we can use the principle of conservation of energy. The potential energy gained by the block can be equated to the kinetic energy of the bullet before the impact. The potential energy (PE) gained by the block at the height of 0.5 meters is given by PE = mgh, where m is the mass of the block, g is the acceleration due to gravity (9.8 m/s2), and h is the height.

PE = 3.0kg × 9.8 m/s2 × 0.5m = 14.7 Joules.

This amount of energy was initially in the form of kinetic energy (KE) of the bullet, which we can calculate using KE = ½mv2.

14.7 J = ½ × 0.75kg × v2

Solving for v, the velocity of the bullet, gives us: v2 = ×2 × 14.7 J / 0.75kg

v2 = 39.2 m2/s2

v = √(39.2 m2/s2)

v ≈ 6.26 m/s

The bullet was traveling at approximately 6.26 m/s before the collision.

Answer 2

The mass of the system block + bullet = 3.75 kg

The initial speed of the system bullet + block is given by

Vf² = Vi² +2*g*d...........


Vf = 0 (Final speed equals zero)

d is the height traveled in the vertical component (0.5m)

Vi = sqrt(-2*(-9.8[m/s²])* (0.5m)) = 3.13 m/s

The momentum is conserved, so

m*V1 (before impact) = M*Vi (after impact)

0.75kg*V1 = 3.75kg*(3.13m/s)

>> V1 = 15.65 [m/s] (Initial speed of the bullet)