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Old Jul 27th 2018, 08:09 AM   #1
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pendulum question

Hi guys,

A bullet of mass 25 grams and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5kg suspended from a chord 3 m long.
How far will the block swing from its position of rest before beginning to return.

so, the bullet has a momentum = mv = 0.025 x 200 = 5 kg m/s

momentum after embedding in the block = (5 + 0.025).u

conservation of momentum gives u = 5/5.025 = 0.995 m/s

so, the pendulum starts to swing at u = 0.995 m/s and will be retarded by the weight of the bullet/block combination. Now according to my book, the angle of the pendulum is calculated by calculating the vertical height of the pendulum as it swings upward to a stop. i.e.

$ v^2 = 2gh $ ............................(1)

where $g = 10m/s^2$ and $h = 3(1 - \cos \theta)$ where $\theta$ is the angle from the vertical.

$ 0.99 = 2. 10. 3(1 - \cos \theta) $

This gives the right answer, but I don't see how it works.
The velocity is initially horizontal, the formula (1) above applies in the vertical direction, where the initial velocity is zero.
Also the equations of motion refer to motion in a straight line, can they be applied to circular motion, or are we just making an approximation?
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Old Jul 27th 2018, 01:29 PM   #2
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Originally Posted by uxinox View Post
Hi guys,

A bullet of mass 25 grams and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5kg suspended from a chord 3 m long.
How far will the block swing from its position of rest before beginning to return.

so, the bullet has a momentum = mv = 0.025 x 200 = 5 kg m/s

momentum after embedding in the block = (5 + 0.025).u

conservation of momentum gives u = 5/5.025 = 0.995 m/s

so, the pendulum starts to swing at u = 0.995 m/s and will be retarded by the weight of the bullet/block combination. Now according to my book, the angle of the pendulum is calculated by calculating the vertical height of the pendulum as it swings upward to a stop. i.e.

$ v^2 = 2gh $ ............................(1)

where $g = 10m/s^2$ and $h = 3(1 - \cos \theta)$ where $\theta$ is the angle from the vertical.

$ 0.99 = 2. 10. 3(1 - \cos \theta) $

This gives the right answer, but I don't see how it works.
The velocity is initially horizontal, the formula (1) above applies in the vertical direction, where the initial velocity is zero.
Also the equations of motion refer to motion in a straight line, can they be applied to circular motion, or are we just making an approximation?
With circular motion its conservation of angular momentum that's conserved. E.g. swing a bob on a string around your head then slowly shorten the string. The force on the bob causes the bob to move in a smaller (instantaneous) circular motion and that increases linear momentum. However the angular momentum remains constant. Of course that's not a closed system but it does work for all systems. But it you have circular motion that's not a closed system since something is causing the particles to accelerate.

For a closed system linear momentum is always conserved. By the way. You can always choose coordinates such that a particle moving in a straight line has angular momentum. Think about that one!
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Old Jul 27th 2018, 02:03 PM   #3
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Originally Posted by uxinox View Post
Hi guys,

A bullet of mass 25 grams and travelling horizontally at a speed of 200 m/s imbeds itself in a wooden block of mass 5kg suspended from a chord 3 m long.
How far will the block swing from its position of rest before beginning to return.

so, the bullet has a momentum = mv = 0.025 x 200 = 5 kg m/s

momentum after embedding in the block = (5 + 0.025).u

conservation of momentum gives u = 5/5.025 = 0.995 m/s

so, the pendulum starts to swing at u = 0.995 m/s and will be retarded by the weight of the bullet/block combination. Now according to my book, the angle of the pendulum is calculated by calculating the vertical height of the pendulum as it swings upward to a stop. i.e.

$ v^2 = 2gh $ ............................(1)

where $g = 10m/s^2$ and $h = 3(1 - \cos \theta)$ where $\theta$ is the angle from the vertical.

$ 0.99 = 2. 10. 3(1 - \cos \theta) $

This gives the right answer, but I don't see how it works.
The velocity is initially horizontal, the formula (1) above applies in the vertical direction, where the initial velocity is zero.
Also the equations of motion refer to motion in a straight line, can they be applied to circular motion, or are we just making an approximation?
Your problem basically refers to a device known as a "ballistic pendulum." You might find this to be interesting.

-Dan
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Old Jul 28th 2018, 01:48 PM   #4
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Many thanks for the link topsquark!

It shows a different method to get

$v^2 = 2gh$

the method shown in my book seemed to get the result by using a horizontal component of the velocity when resolving vertically. It really had me confused.
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