Solve Ballistic Pendulum Height Problem

AI Thread Summary
A ballistic pendulum problem involves a bullet striking a block and determining the height the block rises after the collision. The bullet's mass is 0.024 kg, and it is fired at 97 m/s, emerging from the block at 48.5 m/s. To solve for the height, conservation of momentum is first applied to find the block's velocity immediately after the bullet passes through. The kinetic energy of the block is then equated to gravitational potential energy to find the height, leading to a final height calculation of approximately 0.01016 m. This multi-step problem requires careful attention to mass and velocity values to arrive at the correct solution.
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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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