Physics Question: Elastic Collisions?

Question

Starting with an initial speed of 5.00 m/s at a height of 0.300 m, a
1.50-kg ball swings downward and strikes a 4.60-kg ball that is at rest, as the drawing shows. (a)
Using the principle of conservation of mechanical energy, find the speed of the 1.50-kg ball just
before impact. (b) Assuming that the collision is elastic, find the velocities (magnitude and
direction) of both balls just after the collision. (c) How high does each ball swing after the
collision, ignoring air resistance?

Please show work! (:

Answer 1

Where's the diagram?

I will assume they are tethered on strings and that the velocity you are asking for is at the bottom of the arc

a) 1/2*m*vi^2+m*g*h=1/2*m*vs^2
m divides out
vs^2=vi^2+2*g*h
=5^2+2*9.81*.3
Take the square root to find the velocity

b) since the collision is elastic
both energy and momentum will be conserved
Energy
1/2*1.5*vs^2=1/2*1.5*v(1.5)^2+1/2*4.6*v(4.6)^2
the 1/2 divides out
1.5*vs^2=
1.5*v(1.5)^2+4.6*v(4.6)^2
Momentum
1.5*vs=
1.5*v(1.5)+4.6*v(4.6)

Let's solve for v(1.5)

1.5*v(1.5)^2=
1.5*vs^2-4.6*v(4.6)^2

v(1.5)^2=vs^2-(4.6/1.5)*v(4.6)^2

From the momentum equation
1.5*vs=
1.5*v(1.5)+4.6*v(4.6)
v(4.6)=(1.5/4.6)*(v(1.5)-vs)
v(1.5)^2=vs^2-(4.6/1.5)*[(1.5/4.6)*(v(1.5)-vs)]^2
c) Use conservation of energy

j

Answer 2

You need to either (a) show the drawing or (b) give some information as to the length of the string the balls are hanging on In order for this question to be resolved.

Answer 3

I found the answer to this question using Wikipedia (link attached). The above solutions are correct and yeild the following eqations: v1' = [ v1 * (m1-m2) + 2*m2*v2 ] / [ m1 + m2 ] v2' = [ v2 * (m2-m1) + 2*m1*v1 ] / [ m1 + m2 ] where: v1' = the after collision velocity of object #1 v1 = the before collision velocity of object #1 m1 = the mass of object #1 (same logic for object #2).