Question

The figure shows a ballistic pendulum used to determine the speed of a bullet. A bullet of mass m travelling at a velocity v hits a block of mass M. After collision, the bullet is embedded in the block. The block is displaced through a height h.

Find in terms of m, M, and h
(a) the speed of the bullet
(b) the ratio of the maximum potential energy of the block to the kinetic energy of the bullet.​

Answer 1

A bullet of mass 25 g is fired horizontally into a ballistic pendulum of mass 5.0 kg and gets embedded in it (figure9.E14). If the center of the pendulum rises by a distance of 10 cm, find the speed of the bullet.

Mass of bullet is m = 25 g = 0.025 kg,

Mass of pendulum is M = 5 kg

Vertical displacement is h = 10 cm = 0.1 m,

Let it strike pendulum with velocity u.

Let final velocity be v

Using conservation of energy,

So, the bullet strikes pendulum with velocity u = 280 m/s.

Step-by-step explanation:

Answer 2

The bullet speed after hitting the block is
`2gh/(m+M)`and the ratio of the block's maximum potential energy to the bullet's kinetic energy is
`(m+M)^2/2gh.`

(a) Speed of the bullet

When the bullet hits the block, it gets embedded in it. This means that the bullet and block move together as a single system after the collision. This is a perfectly inelastic collision.

Momentum is conserved in a perfectly inelastic collision, so we can write:

`m * v + M * V = (m + M) * V'`

where:

* m is the mass of the bullet

* v is the initial velocity of the bullet

* M is the mass of the block

* V is the initial velocity of the block (which is zero)

* V' is the final velocity of the bullet-block system

After the collision, the bullet-block system swings up to a height h. This means that its kinetic energy has been converted into potential energy. We can use this to write another equation:

`(1/2) * (m + M) * V'^2 = (m + M) * g * h`

where g is the acceleration due to gravity.

Now we have two equations, and we can solve for our two unknowns, v and V'.

Dividing the first equation by the second equation, we get:

`(v + MV') / V' = 2gh / V'^2`

Multiplying both sides of the equation by
`V'^2,` we get:

`v + MV' = 2gh`

Solving for v, we get:

`v = 2gh / (M + m)`

(b) Ratio of the maximum potential energy of the block to the kinetic energy of the bullet

The maximum potential energy of the block is reached when it is at its highest point. This is when the kinetic energy of the bullet-block system has been completely converted into potential energy.

The maximum potential energy of the block is given by:

`(m + M) * g * h`

The kinetic energy of the bullet is given by:

`(1/2) * m * v^2`

Therefore, the ratio of the maximum potential energy of the block to the kinetic energy of the bullet is:

`(m + M) * g * h / (1/2) * m * v^2 = 2gh / v^2`

Substituting the expression for v that we found in part (a), we get:

`2gh / v^2 = 2gh / (2gh / (M + m))^2 = (M + m)^2 / 2gh`

Therefore, the ratio of the maximum potential energy of the block to the kinetic energy of the bullet is:

`(M + m)^2 / 2gh`