Question

An 8-kg block is hanging on a string when a .01 kg bullet is shot into it and gets stuck inside it from below. If the bullet and block rise to a height of 1.6m, what velocity did the bullet enter the block?

Answer 1

The velocity of the bullet just before it hits the block is approximately √2gh, where g is the acceleration due to gravity (approximately 9.8 m/s²) and h is the height (1.6 m).

Step-by-step explanation:

This problem involves the principle of conservation of energy. The initial potential energy of the block and the bullet is converted into the final potential energy when they reach a certain height.

The kinetic energy of the bullet just before it hits the block is also converted into potential energy.

The potential energy (PE) gained is given by the formula:

PE=mgh

Where:

m is the mass (combined mass of the block and bullet),

g is the acceleration due to gravity (approximately 9.8 m/s²),

h is the height.

The initial potential energy is the potential energy of the block alone:

PEinitial = mblockgh

The final potential energy is the potential energy of the combined mass of the block and the bullet:

PEfinal = (mblock + mbullet)gh

The kinetic energy of the bullet just before impact is given by:

KE= 1/2 mbulletv2

Where:

v is the velocity of the bullet.

Since energy is conserved, the initial potential energy is equal to the sum of the final potential energy and the kinetic energy:

mblockgh=(mblock + mbullet )gh+ 1/2mbulletv2

Now, you can substitute the known values into the equation and solve for

8×9.8×1.6=(8+0.01)×9.8×1.6+ 1/2×0.01×v2

Solve for v to find the velocity of the bullet just before it hits the block.