Solving Clay Collision: Final Speed & Swing Height

[lemonlimesoap]

Homework Statement

A 2kg. lump of clay traveling at a speed of 20m/s collides with and sticks to a 5kg. lump of clay initially at rest and hanging from a string. Find the final speed of the new lump of clay. How high will the clay swing after the collision?

(initial)
m1=2kg.
v1=20m/s

(final)
m2=2kg.+5kg.=7kg.
v2=?

Homework Equations

KEi=PEf

KEi=1/2(m1)(v1)^2

PEf=(m2)(g)(h)

v=(2gh)^1/2

P=mv - I'm pretty sure you don't need this one

The Attempt at a Solution

KEi=PEf
(1/2)(2kg.)(20m/s)^2=(7kg.)(9.81m/s^2)(h)
solve for h=?
h=5.825m

solve for v2=?
v2=(2gh)^1/2
v2=10.69m/s

According to the answer given by my teacher h=1.66m if this is correct please show me how.

 

[lemonlimesoap]

posted in the wrong section by accident, admins can move it if they want.

 

[HallsofIvy]

Homework Statement

A 2kg. lump of clay traveling at a speed of 20m/s collides with and sticks to a 5kg. lump of clay initially at rest and hanging from a string. Find the final speed of the new lump of clay. How high will the clay swing after the collision?

(initial)
m1=2kg.
v1=20m/s

(final)
m2=2kg.+5kg.=7kg.
v2=?

Homework Equations

KEi=PEf

KEi=1/2(m1)(v1)^2

PEf=(m2)(g)(h)

v=(2gh)^1/2

P=mv - I'm pretty sure you don't need this one

The Attempt at a Solution

KEi=PEf
(1/2)(2kg.)(20m/s)^2=(7kg.)(9.81m/s^2)(h)

Here is your error. In an inelastic collision, energy is NOT conserved. Use conservation of momentum to find the velocity of both lumps of clay immediately after the collision and use that rather than "20 m/s".

solve for h=?
h=5.825m

solve for v2=?
v2=(2gh)^1/2
v2=10.69m/s

According to the answer given by my teacher h=1.66m if this is correct please show me how.