# Perfectly elastic collision

by sktgurl930
Tags: collision, elastic, perfectly
 Share this thread:
 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
HW Helper
P: 5,341
 Quote by sktgurl930 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)
P: 21
 Quote by LowlyPion 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

HW Helper
P: 5,341
Perfectly elastic collision

 Quote by sktgurl930 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.

 Related Discussions Introductory Physics Homework 15 Introductory Physics Homework 5 Introductory Physics Homework 7 Introductory Physics Homework 2 Introductory Physics Homework 10