How Do You Calculate Initial Velocity in an Elastic Collision Problem?

[David Lee]

I need help!

Homework Statement

A 44.0 g ball is fired horizontally with initial speed vi toward a 110 g ball that is hanging motionless from a 1.10 m-long string. The balls undergo a head-on, perfectly elastic collision, after which the 110 ball swings out to a maximum angle = 52.0.

What was vi ?

Homework Equations

Conservation of momentum:
m1vi1 + m2vi2 = m1vf1 + m2vf2

Conservation of energy:
1/2m1(vi1^2) = m2gy

The Attempt at a Solution

I tried to solve these problem with those 2 equations, but it still doesn't work. I compeletly massed up with these concepts. Can anyone help me with this problem with exact answer?
Thank you

 

[mgb_phys]

You almost have the energy conservation right.

ke of first ball -> ke of second ball -> potential energy of second ball.
you know m1, m2, and h so just find v1

 

[David Lee]

I do not know how to combine those two equations. Can you through specific steps by using numbers?

 

[David Lee]

You almost have the energy conservation right.

ke of first ball -> ke of second ball -> potential energy of second ball.
you know m1, m2, and h so just find v1

I do not know how to combine those two equations. Can you through specific steps by using numbers?

 

[mgb_phys]

ke of first ball = 1/2 m1 v^2
pe of 2n ball = m2 g h

Just set them equal
1/2 * m1 * v1^2 = m2 * g * h
1/2 * 0.044 * v1^2 = 0.110 * 9.8 * h

You need to do a bit of trig to get h, then it's simple

 

[David Lee]

ke of first ball = 1/2 m1 v^2
pe of 2n ball = m2 g h

Just set them equal
1/2 * m1 * v1^2 = m2 * g * h
1/2 * 0.044 * v1^2 = 0.110 * 9.8 * h

You need to do a bit of trig to get h, then it's simple

okay, thank your so far. I got the h as 0.110cos53, but is it right? or h is (0.110 - 0.110cos53 )?
which one is right?

 

[kuruman]

ke of first ball = 1/2 m1 v^2
pe of 2n ball = m2 g h

Just set them equal
1/2 * m1 * v1^2 = m2 * g * h
1/2 * 0.044 * v1^2 = 0.110 * 9.8 * h

You need to do a bit of trig to get h, then it's simple

There is no wording in the problem that says that the 44 g is at rest after the collision. It seems to me that the energy conservation equation is missing the kinetic energy of the 44 g ball after the collision. One needs to solve the momentum conservation equation for the velocity of the 44 g mass after the collision and substitute in the modified energy conservation equation.