perfectly elastic collision


by sktgurl930
Tags: collision, elastic, perfectly
sktgurl930
sktgurl930 is offline
#1
Nov3-08, 12:08 PM
P: 21
1. The problem statement, all variables and given/known data
A 15 g ball is fired horizontally with speed v0 toward a 117 g ball hanging motionless from a 1.53-m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 117 g ball swings out to a maximum angle θmax=53. What was v0?

2. Relevant equations
h=L(L-cos(theta)
V=Square root of (m*g*h/(.5*M) little m is ball moving and M is the ball not moving
(v2x)f=2m1/m1+m2 (V1x)i



3. The attempt at a solution
Im not sure if im using the right equations and whether the angle given is the one we use in the equation, or if we have to subtract it from 90 first
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LowlyPion
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#2
Nov3-08, 12:41 PM
HW Helper
P: 5,346
Quote Quote by sktgurl930 View Post
1. The problem statement, all variables and given/known data
A 15 g ball is fired horizontally with speed v0 toward a 117 g ball hanging motionless from a 1.53-m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 117 g ball swings out to a maximum angle θmax=53. What was v0?

2. Relevant equations
h=L(L-cos(theta)
V=Square root of (m*g*h/(.5*M) little m is ball moving and M is the ball not moving
(v2x)f=2m1/m1+m2 (V1x)i

3. The attempt at a solution
Im not sure if im using the right equations and whether the angle given is the one we use in the equation, or if we have to subtract it from 90 first
A couple of things.

Initial KE is 1/2*m*v2

The second is that the angle because it is hanging is with the vertical. Hence the height should be given by the Cosθ times the string length subtracted from the length. (h = length - projection of the string to the vertical.)

So ... 1/2*m1*v2 = m2*g*(L - L*cos53)

v2 = 2*(m2/m1)*g*(L-L*Cos53)
sktgurl930
sktgurl930 is offline
#3
Nov3-08, 01:07 PM
P: 21
Quote Quote by LowlyPion View Post
A couple of things.

Initial KE is 1/2*m*v2

The second is that the angle because it is hanging is with the vertical. Hence the height should be given by the Cosθ times the string length subtracted from the length. (h = length - projection of the string to the vertical.)

So ... 1/2*m1*v2 = m2*g*(L - L*cos53)

v2 = 2*(m2/m1)*g*(L-L*Cos53)
so my mass 1 is the ball not moving right??
is this the equation i use to get the answer or do i have to plug it into another one

LowlyPion
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#4
Nov3-08, 02:09 PM
HW Helper
P: 5,346

perfectly elastic collision


Quote Quote by sktgurl930 View Post
so my mass 1 is the ball not moving right??
is this the equation i use to get the answer or do i have to plug it into another one
Sorry my mistake. The m1 I wrote on the left should be m2 - the stationary ball. We are trying to calculate the velocity of m2 AFTER impact. The m's should cancel there. Sorry for my typo which I then proceeded to run with.

With that final velocity for the stationary ball after impact and the usual equations for conservation of energy and momentum, you should now have 2 equations and 2 unknowns, one of which is the Vo that they ask for.


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