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snow_tiger [21]
3 years ago
15

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
Physics
1 answer:
pav-90 [236]3 years ago
8 0

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!

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