Question

A 11.0 g wad of sticky clay is hurled horizontally at a 120 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?
m/s

Answer 1

The masses are given as 11 g and 120 g, by converting them into kg we get,0.011 and 0.120 kg respectively. The total mass is given as,
(11.0 + 120)/1000 kg, or 0.131 kg.
There is only one force producing a net acceleration when the clay and block begins sliding. Let us assume direction of the block's motion be positive, so that friction is acting in the negative direction.

which means

-f_k = ma (f_k = force of kinetic friction, and is different from f, force. m=mass and a= acceleration)
-f_k = ( = coefficient of friction)
n = mg as there is no motion in the vertical direction
-fk = -mg = ma
-mg = ma
-g = a
-(0.650)(9.80) = a
a = -6.37 m/s²

So there will be an acceleration of -6.37 m/s^2, when the block starts to move. we can use simple kinematic equations in one direction to find out what the initial speed of the object in order to move 7.50 m while deaccelerating.

v_f² = v_i² + 2ad

where v_f = final velocity 0 since the block stops at the end

v_i = initial velocity we need to find it out.

a = acceleration ,we find it as ,-6.37m/s²

d = displacement ,given as 7.50 m

By substituting the values we get
0² = v_i² + 2(-6.37)(7.50)

0 = vi² - 95.55
v_i² = 95.55
=> v_i = 9.77 m/s

Initial speed of the clay block system was 9.77 m/s.

Now use the conservation of momentum.

initial momentum = final momentum

From the conservation theorm,

clay's momentum + block's momentum = clay-block's momentum
block is at rest initially ,so momentum of block=0
clay's momentum = clay-block's momentum
0.011*clay's horizontal speed = 0.131*(9.77)
0.011v = 1.28
v = 116.4 m/s

The speed of the clay immediately before impact is ,v=116.4 m/s