How can I determine the speed of a bullet fired into a block of wood?

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To determine the speed of a bullet fired into a block of wood, the conservation of momentum principle is applied. Initially, the bullet's mass is 5 grams, and the block's mass is 10 kilograms, with the block moving at 0.5 m/s after the collision. The correct calculation shows that the bullet's speed is 100.5 m/s, not the previously calculated 2.5 m/s. The error was in the total mass used in the momentum equation, which should be 1005 grams instead of 15 grams. Understanding these calculations is crucial for accurately determining bullet speed in such scenarios.
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Bullet speed- Please Help!

I am not to great at physics and I was hoping someone could point me where I am going wrong in this problem! :) My answer I came up with is 2.5 m/s but that isn't one of my options so I know something has to be off.


5. To find the speed with which a gun fires bullets, a bullet of mass 5 gm is fired horizontally into a 10 Kg block of wood resting on a frictionless table. The bullet gets embedded in the block and the block is found to have a velocity of 0.5 m/s after collision. What is the speed of the bullet?

Bullet Speed= 2.5 m/s

Mv1i + 0 = (m +M)V
Mv1i + 0 = (5gm +10 kg) .5 m/s
m1 u1 + m2 u2 = (m1 + m2) V
(5gm u1) + (10kg *0m/s) = (5gm + 10kg) .5m/s
(5gm u1) + (0) =7.5
u1 = 2.5 m/s
 
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Think of the conservaton off momentum. Remember the bullet is embeded in the block.

added by edit: Never mind. I didn't look at your work well enough. You've got the idea. But (5gm + 10kg)=1005gm not 15gm.

regards
 
Last edited:
So... :)

So would this be more on track:


Mv1i + 0 = (m +M)V
Mv1i + 0 = (5gm +10 kg) .5 m/s
m1 u1 + m2 u2 = (m1 + m2) V
(5gm u1) + (10kg *0m/s) = (5gm + 10kg) .5m/s
(5gm u1) + (10kg *0m/s) = (1005gm) .5m/s
(5gm u1) + (0) = 502.5
u1 = 100.5 m/s


Bullet Speed= 100.5 m/s


Thank you so much for your help! :)

Emma
 
Any time. And welcome to the forums.
 
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