Question

A small wooden block with mass m1 is suspended from the lower end of a light cord that is l long. The block is initially at rest. A bullet with mass m2 is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of h, the tension in the cord is T0. What was the initial speed v0 of the bullet

Answer 1

Answer:

Explanation:

The tension of the cord is not important, what matters is the height the block has risen.

When it has risen to its maximum it will have a speed of 0, and because of that it will have a kinetic energy of 0. However it will have a higher potential energy than it had at the beginning. The difference in potential energy will be:

$\Delta Ep = (m1 + m2) * g * h$

The energy to rise the block with the bullet embedded into it came from the kinetic energy of the bullet.

$Ec = \frac{1}{2} * m2 * v0^2$

These two energies are equal because all the kinetic energy the bullet had was transformed into potential energy. Therefore:

$(m1 + m2) * g * h = \frac{1}{2} * m2 * v0^2$

Rearranging:

$V0 = \sqrt{\frac{2*(m1 + m2) * g *}{m2}}$