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!