Solve Ballistic Pendulum Height Problem

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Homework Help Overview

The problem involves a ballistic pendulum scenario where a bullet strikes a block, and the objective is to determine the height the block rises after the collision. The context includes concepts from momentum conservation and kinetic energy.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using conservation of momentum to find the velocity of the block after the bullet passes through. There are attempts to apply kinetic energy principles to find the height the block rises.

Discussion Status

Several participants are exploring different methods to calculate the block's height, with some questioning the energy lost during the collision. There are indications of confusion regarding the correct application of formulas and values, and multiple interpretations of the problem setup are being examined.

Contextual Notes

Participants note discrepancies in their calculations and the values used, including the mass of the bullet and the velocities involved. There is an ongoing exploration of assumptions related to energy conservation and momentum transfer during the collision.

shadowice
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[solved]Ballistic pendulum

Homework Statement


A bullet of mass m = 0.024 kg is fired with a speed of vi = 97.00 m/s and hits a block of mass M = 2.61 kg supported by two massless strings. The bullet emerges from the right side of the block with a speed of vf = 48.50 m/s. Find the height to which the block rises.

m1= 0.024 kg
m2= 2.61 kg
vi = 97 m/s
vf = 48.5 m/s
h = unknown

prob06-1.gif


Homework Equations


Before+JA collision
m1v1+ m2V2 = P
Ki = 1/2m1v1^2+1/2m2vf^2
1/2mv^2= mgh

The Attempt at a Solution


I know i need to use con of momentum before and just after bullet contact to find v. But that's where I am having problems at, I am not really sure how to set it up to find v

I had been doing 1/2m1v1^2+1/2m2vf^2 = 3182.6 then trying to set that = mgh but it gives me 120.

If i can find that v i know id need 1/2*m2*V = m2*g*h
then solve from h there if I am correct if someone could give me nudge into right direction would be great
 
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I don't notice where you take into account the energy lost during the collision.
 
ok so would i do 1/2mvi^2 = 1/2 mVm2^2
vm2= velocity of block just after bullet passes threw
1/2*.024*97^2= 1/2*2.61*vm2^2

vm2 = 9.3016

1/2mvm2^2= mgh
1/2*9.3016^2 = 9.8h (m's cancel out)

i got h = 4.414 m but tells me that's wrong
 
this is a complicated multi-step problem.

first, use conservation of momentum to find the velocity of the block when the bullet is embedded (.884 m/s). Then, use conservation of momentum again using equation mass (bullet plus block)*velocity (bullet plus block)= Mbullet*Vfbullet+ Mblock*Vfblock

subtract over the bullet final and divide by the mass of the block to obtain the velocity of the block after the bullet exits. Then, use conservation of kinetic energy to determine how high the block goes. 1/2mv^2=mgy Solve for y, and that's the answer. I got .1 m for the answer so if that's wrong I could be off.
 
that didnt work but where are you getting the .1 when i do what you typed out i get
(.024+2.61)*.884 = 2.61*48.5+2.61*Vb
2.328456 = 126.585 + 2.61*VB
-124.26 = 2.61*vb
47.6 m/s = vb

plugging that # in for v i get no where near .1, so you could be close but it wanted atleast 4 sig digits and adding 3 zeros wasnt it
 
be careful, in your equation you entered 48.5 m/s as the block's velocity after impact. You should be solving for this variable, so insert 48.5m/s as the bullet's velocity. There was an error on my part; the height should be .0099 m
 
thnx i put in wrong mass for bullet was the problem got it now was .01016
 

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