# physics

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=2+5=7kg.
v2=?

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

My friend was very adamant in telling me that the h=1.66m. Either I messed up in the equation I used or he did.

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1. You are very wrong.

first, v2: conservation of momentum applies 2*20=(2+5)V or V= 2/7 *20 m/s

Now, the KE this has is 1/2 m v^2 or
1/2 (7)(4/49 * 400) work that out.
this is now equal to mgh or 7*9.81*h
solve for h.

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bobpursley

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