Question

A pendulum consists of a 2.0-kg block hanging on a 1.5-m length string. A 10-g bullet moving with a horizontal velocity of 900 m/s strikes, passes through, and emerges from the block (initially at rest) with a horizontal velocity of 300 m/s. To what maximum height above its initial position will the block swing

Answer 1

Answer:

The maximum height above its initial position is:

$h_{max}=1.53\: m$

Explanation:

Using momentum conservation:

$m_{b}v_{ib}=m_{B}v_{fB}+m_{b}v_{fb}$ (1)

Where:

• m(b) is the mass of the bullet
• m(B) is the mass of the block
• v(ib) is the initial velocity of the bullet
• v(fb) is the final velocity of the bullet
• v(fB) is the final velocity of the block

Let's find v(fb) using equation (1)

$m_{b}(v_{ib}-v_{fb})=m_{B}v_{fB}$

$v_{fB}=\frac{m_{b}(v_{ib}-v_{fb})}{m_{B}}$

$v_{fB}=\frac{0.1(900-300)}{2}$

$v_{fB}=30\: m/s$

We need to find the maximum height, it means that all kinetic energy converts into gravitational potential energy.

$\frac{1}{2}m_{B}v_{fB}=m_{B}gh_{max}$

$h_{max}=\frac{1}{2g}v_{fB}$

$h_{max}=\frac{1}{2(9.81)}30$

$h_{max}=1.53\: m$

I hope it helps you!