Elastic Collision between projectile and a pendulum, what's the initial speed?

AI Thread Summary
A 29.0 g ball is fired horizontally at a 100 g pendulum ball, leading to a perfectly elastic collision where the pendulum swings to a maximum angle of 50.0°. The conservation of momentum and energy equations are applied to find the initial speed of the projectile. Calculations indicate that the final velocity of the pendulum (vp) is approximately 2.775 m/s, and the initial speed (v0) is determined to be around 6.17 m/s. The discussion highlights the importance of correctly applying mass ratios and conservation principles in elastic collisions. The final answer for the initial speed of the projectile is confirmed to be 6.17 m/s.
Fromaginator
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Homework Statement


A 29.0 g ball is fired horizontally with initial speed v0 toward a 100 g ball that is hanging motionless from a 1.10 m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 100 g ball swings out to a maximum angle θmax = 50.0°
see image for question

Homework Equations


Ki + Ui = Kf + Uf
vp = √[2*g*L*(1 - cosθ)]
mb*v0 = mp*vp + mb*vb



The Attempt at a Solution


I suspect you have to use the mass ratio and it's relation to the ratio of momentum or kinetic energy, but I have been unsuccessful in getting the correct answer. The tried answers are availible along with the question in the image.
 

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I got a different answer. But I can't just tell you - that would be spoiling your learning. If you show your calculation, I can suggest improvements.
 
I think the problem with my calculations was that I used a mass ratio which I think only applies to inelastic but here's what I did
0.5*mp*vp² = mp*g*L*(1 - cosθ)

vp = √[2*g*L*(1 - cosθ)]

0.5*mb*v0² = 0.5*mb*vb² + 0.5*mp*vp²
mb*v0² = mb*vb² + mp*vp²
mb*vb² = mb*v0² - mp*vp²
vb = √[v0² - (mp/mb)*vp²]

v0 = √[v0² - (mp/mb)*vp²] + (mb/mp)*vp


mb = 27 g
mp = 100 g
vp = √[2*g*L*(1 - cosθ)]; L = 1.1 m, θ = 50º

vp = 2.775 m/s

mp/mb = 3.70

v0 = √[v0² - (3.7)*7.701] + (3.7)*2.775
v0 = √[v0² - 28.5] + 10.27
v0 - 10.27 = √[v0² - 28.5]
v0² - 20.54*v0 + 105.4 = v0² - 28.5
- 20.54*v0 + 105.4 = - 28.5

v0 = 6.52 m/s

I'm pretty sure the vp(final velocity of the pendulum is right)
 
Fromaginator said:
v0 = √[v0² - (mp/mb)*vp²] + (mb/mp)*vp

This came from the conservation of momentum equation, but you made a mistake here. If you write out the full momentum equation, you'll see what it should be.
 
I got it!
vp=[2mb/(mb+mp)]*vb
0.5*mvbf^2+mghf=0.5*mvbi^2+mghi
gL(1-costheta)=).5(Vo^2)(3364/16641)
Vo=6.17m/s

although when I actually did it there were a few more intermeadiate steps but I didn't feel it necessary to type out the algebra since square roots and fractions are hard to type.
 
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