What is the Initial Velocity v0 in a Perfectly Elastic Collision?

AI Thread Summary
The discussion focuses on calculating the initial velocity (v0) of a 15 g ball before a perfectly elastic collision with a 117 g ball. Key equations involve conservation of momentum and energy, specifically using the height derived from the angle θmax=53° to find the potential energy of the stationary ball. The height is calculated as h = L - L*cos(θ), which is crucial for determining the final velocity of the stationary ball post-collision. Participants clarify that the mass assignments in the equations need to be consistent, particularly identifying which ball is moving and which is stationary. Ultimately, the correct application of these principles will yield the desired initial velocity v0.
sktgurl930
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Homework Statement


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?

Homework 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



The Attempt at a Solution


Im not sure if I am 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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sktgurl930 said:

Homework Statement


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. Homework 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

The Attempt at a Solution


Im not sure if I am 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)
 
LowlyPion said:
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
 
sktgurl930 said:
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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