Answer: Ballistic Pendulum: Bullet Collision & Max Height

AI Thread Summary
A rifle bullet of mass 2.5 g traveling at 243 m/s collides with a pendulum of mass 237.5 g, initially at rest, resulting in a combined momentum of 0.6075 kg m/s and an initial velocity of 2.53125 m/s for the pendulum. The discussion focuses on determining the maximum height the bullet-pendulum combination reaches after the collision, with participants struggling to find the correct angle and height calculations. The energy conservation approach is debated, with emphasis on the need to apply momentum conservation principles instead. The correct formula for the final height involves using the initial velocity derived from the bullet's momentum, leading to a reevaluation of the height calculation. The conversation highlights the complexities of applying both momentum and energy conservation in collision scenarios.
nahya
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A rifle bullet of mass m = 2.5 g traveling at vb = 243 m/s collides with and embeds itself in a pendulum of mass M = 237.5 g, initially at rest and suspended vertically with massless strings of length L = 2 m.
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i first converted the masses into kilograms.
i found out that net momentum = 0.6075 kg m/s, for the bullet-pendulum mass of 0.24kb. that means that the pendulum moves at first with v = 2.53125.

now... where do i go from here?
i don't even know the angle at which the bullet-pendulum combination reaches the max height. if i knew, it would be 2*tan(theta).

so i guess the challenge is to find the angle at which the thing goes at its max height...? bleh.. I'm lost :cry:

edit:

by the energy equation, i guess...
for M = total mass, 1/2*Mv(f)^2 + Mgy(f) = 1/2*Mv(i)^2 + Mgy(i)
y(i) is zero, so the last term cancels out.
v(f) is zero, so the first term cancels out.
Mgy(f) = 1/2*Mv(i)^2
y(f) = Mv(i)^2 / (2Mg)
i get 0.6538, which is not the answer...
 
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nahya said:
A rifle bullet of mass m = 2.5 g traveling at vb = 243 m/s collides with and embeds itself in a pendulum of mass M = 237.5 g, initially at rest and suspended vertically with massless strings of length L = 2 m.
---
i first converted the masses into kilograms.
i found out that net momentum = 0.6075 kg m/s, for the bullet-pendulum mass of 0.24kb. that means that the pendulum moves at first with v = 2.53125.

now... where do i go from here?
i don't even know the angle at which the bullet-pendulum combination reaches the max height. if i knew, it would be 2*tan(theta).

so i guess the challenge is to find the angle at which the thing goes at its max height...? bleh.. I'm lost :cry:

edit:

by the energy equation, i guess...
for M = total mass, 1/2*Mv(f)^2 + Mgy(f) = 1/2*Mv(i)^2 + Mgy(i)
y(i) is zero, so the last term cancels out.
v(f) is zero, so the first term cancels out.
Mgy(f) = 1/2*Mv(i)^2
y(f) = Mv(i)^2 / (2Mg)
i get 0.6538, which is not the answer...
What is the question? Do you want to find the angle? If so, you have worked out the right expression for y(f) and from that you can find the angle(see correction below). Do a drawing.
AM
 
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i'm trying to find the maximum height that it reaches.
i keep getting the same answer for y(f), and it is, apparently, incorrect.
 
nahya said:
i'm trying to find the maximum height that it reaches.
i keep getting the same answer for y(f), and it is, apparently, incorrect.
ok. Your expression for y(f) is not right. Energy is not conserved. The momentum of the bullet before the impact is equal to the momentum of the block + bullet afterward. The velocity immediately after the impact has to be worked out from the bullet momentum (mv_b).

So:

v_i = \frac{m}{m+M}v_b

and:

\frac{1}{2}(M+m)v_i^2 = (M+m)gy_f

AM
 
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