- #1

scruffles

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## Homework Statement

A ball of clay collides with a block that sits on a frictional surface. If the collision proceeds in 1 dimension and the clay sticks to the block, how far does the block/clay system slide before coming to a stop? The clay has mass m = 50 kg, the block as mass M = 100 kg, the initial velocity of the clay is 20 m/s, and the coefficient of kinetic friction between the block and the surface is 0.2.

## Homework Equations

m1v1+m2v2=(m1+m2)vf

v_ave=(vi+vf)/2

## The Attempt at a Solution

50kg(20 m/s) + 100kg(0 m/s) = (50kg+100kg)vf

= 1000 kgm/s=150kgvf

= 6.66m/s = vf (speed of both objects after collision)

Rate of deceleration :

.2 x 9.8m/s^2 = 1.96 m/s^2

The velocity is 6.66 m / s, so the time it takes to stop is:

6.66 / 1.96 ~= 3.40 s

Average speed with linear deceleration is 1/2 initial speed or:

6.66 * 1/2 = 3.33 m/s

So the distance traveled is average speed times the time it took to stop:

3.33 m/s * 3.40 s ~ 11.33 m.

Did I do this right or wrong?

Thanks in advance!