Putty Physics Problem?

Question

A 15 kg block is at rest on a level floor. A 390 g glob of putty is thrown at the block such that it travels horizontally, hits the block, and sticks to it. The block and putty slide 15 cm along the floor. If the coefficient of sliding friction is 0.4, what is the initial speed of the putty?

Answer 1

Hmmm
arbiter007 is on right track

First conservation of momentum

m1V1=(m1+m2)V2

m1- mass of the putty
m2 –mass of the block
V1-initial velocity of the putty
V2- velocity of the putty and the block

Then energy equation
Kinetic energy after impact = work done against friction

.5 (V2)^2(m1+m2)=FS=u (m1+m2) g S
Where
g- gravitational acceleration
S- distance the block and putty traveled after impact
u - coefficient of friction

from last equation
V2=sqrt(2 u g S)

From the first

V1=(m1+m2)V2/m1=[(m1+m2)/m1] sqrt(2 u g S)
V1=[(.390+15)/.390]sqrt(2 (.4) 9.81 .15)=
V1=[39.5]sqrt(1.1772)=42.9m/s

I hope that helps

Answer 2

Force times distance equals work. The work is equal to the initial kinetic energy of the combined mass, which gives you the initial velocity. Since this is an inelastic collision, conservation of momentum will allow you to solve for the initial momentum of the putty, hence the velocity thereof.