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October 19th, 2015, 01:49 AM   #1
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Another Linear momentum Question

A 22g bullet strikes and becomes embedded in a 1.35kg block of wood placed on a horizontal surface just in front of the gun.
If the co-efficient of kinetic friction is 0.28, and the impact drives the block a distance of 8.5m before it comes to rest, what was the muzzle speed of the bullet?

$\displaystyle F_{fr} = F_N \mu_k$

$\displaystyle m\cdot a = m\cdot g \cdot\mu_k$

$\displaystyle a = g\cdot \mu_k$

$\displaystyle a = 9.8(0.29) = -2.744ms^{-2}$

$\displaystyle v^2 = v_0^2 + 2ad$, final speed after collision is 0m/s

$\displaystyle v_0 = \sqrt{-2ad}$

$\displaystyle v_0 = \sqrt{[-2]\cdot [-2.744]\cdot[8.5]}$

$\displaystyle v_0 = 6.83ms^{-1} = v'$

$\displaystyle p = p'$

$\displaystyle m_1v_1 = (m_1 + m_2)v'$

$\displaystyle v_1 = \frac{(0.022 + 1.35)\cdot 6.83}{0.022}$

muzzle speed = $\displaystyle v_1 = 397.88 = 398ms^{-1}$, Is this correct?? Thank you in advance.
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October 19th, 2015, 03:34 PM   #2
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check your arithmetic for the final calculation of $v_1$ ... I get approx 426 m/s
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October 20th, 2015, 08:55 PM   #3
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Thanks skeeter...

recalculated and $\displaystyle v_1 = 487ms^{-1}$
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October 21st, 2015, 07:04 AM   #4
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Originally Posted by hyperbola View Post
Thanks skeeter...

recalculated and $\displaystyle v_1 = 487ms^{-1}$
huh?
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October 21st, 2015, 07:12 AM   #5
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apologies...

My calculator gives 487. I've tried another calculator (CASIO) and it now gives 425.9. Just to make sure, I used an online calculator and yes it is 425.9.

Don't know why my calculator is giving 487.

Last edited by hyperbola; October 21st, 2015 at 07:14 AM.
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October 22nd, 2015, 06:11 AM   #6
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You get 487 m/s if you accidentally type 0.22 instead of 0.022 in the bracket on the numerator.
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