Upcoming Physics Olympiad Test

AI Thread Summary
A bullet of mass m collides inelastically with a pendulum of mass M, and the goal is to determine the minimum initial velocity v0 for the combined system to complete a vertical loop. The key equations involve momentum conservation and energy conservation, specifically relating kinetic and potential energy. The initial solution mistakenly calculated the velocity needed to just reach the top of the loop instead of the minimum required to complete it. The correct approach involves recognizing that the bob must maintain a non-zero velocity at the top of the loop, leading to a revised equation for v0. Ultimately, the correct formula for v0 is derived as ((m+M) sqrt(5Lg)) / m.
physics517
Messages
7
Reaction score
0

Homework Statement



Abullet of mass m strikes a pendulum of mass M suspended from a pivot by a string of length L with a horizontal velocity v0. The collision is perfectly INelastic and the bullet sticks to the bob. Find the minimum velocity v0 such that the bob (with the bullet inside) completes a sircular vertical loop


Homework Equations



p=mv

KE=1/2 mv^2

PE=mgh

The Attempt at a Solution



mvo=(m+M)vf

mg(2L)=1/2 m vf^2 --------- This is the Vf needed to comlete one whole loop (also i think this is where I am missing something)

vf=2 sqrt(Lg)

and so the answer i got was vo= ((m+M)2 sqrt(Lg)) / m

but the ACTUAL asnwer is vo= ((m+M) sqrt(5Lg)) / m
 
Last edited:
Physics news on Phys.org
What equations are relevant? Where's your attempt at a solution?

By the way, your post title is pretty unspecific about the actual problem content.
 
physics517 said:
mg(2L)=1/2 m vf^2 --------- This is the Vf needed to comlete one whole loop (also i think this is where I am missing something)
Indeed. That is NOT the velocity needed to complete one loop, that is the minimum velocity required for the bob (and bullet) to just reach the top of the circular loop. For the bob and bullet to complete a circular loop, it cannot have zero velocity at the top of the loop! There is a minimum velocity that it must possesses - think back to your concepts of circular motion.
 
Lol I shouldve drew a diagram.

so up top its F+mg=mv^2/r and the F (tension) cancels out since we're trying to find minimum speed. so v=sqrt(gr) and if we go back to my original work equation

it should be 1/2 m vf^2 = mgh + 1/2 m vtop ^2

then you find the Vf and then plug that into the momentum equation

Thanks so much guys!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top