An 8.00-kg block of wood sits at the edge of a frictionless table, 2.20 m above the floor. A 0.500-kg blob of clay slides along the length of the table with a speed of 24.0 m/s, strikes the block of wood and sticks to it. The combined object leaves the edge of the table and travels to the floor. What horizontal distance has the combined object traveled when it reaches the floor?

Given Data:

The speed of clay before the strike is $\mathrm{u}=24\hspace{0.33em}\mathrm{m}/\mathrm{s}$

The height of the woodblock above the floor is $\mathrm{h}=2.20\hspace{0.33em}\mathrm{m}$

The mass of clay is: $\mathrm{m}=0.500\mathrm{kg}$

The mass of the block of wood is: $\mathrm{M}=8\hspace{0.33em}\mathrm{kg}$

The conservation of momentum before and after a collision of Clay and wood block is used to find the speed of the combined object after the strike. The horizontal distance is found by the product of time for the combined object to reach the ground and the speed of the combined object.

The time taken by the combined object to reach the floor is given as:

$t=\sqrt{\frac{2h}{g}}$

Here, g is the gravitational acceleration and its value is $9.8\hspace{0.33em}\mathrm{m}/{\mathrm{s}}^2$.

Substitute all the values in the above equation.

$$\begin{gathered} t=\sqrt{\frac{2(2.20\hspace{0.33em}m}{9.8\hspace{0.33em}m/s^2}} \\ t=0.670\hspace{0.33em}s \end{gathered}$$

Apply the conservation of momentum to find the speed of combined object.

$\mathrm{mu}=(\mathrm{m}+\mathrm{M})\mathrm{v}$

v is the speed of combined object.

Substitute all the values in the above equation.

$$\begin{gathered} \left(0.500\mathrm{kg}\right)\left(24\mathrm{m}/\mathrm{s}\right)=\left(0.500\mathrm{kg}+8\mathrm{kg}\right)\mathrm{V} \\ =1.412\mathrm{m}/\mathrm{s} \end{gathered}$$

The horizontal distance travelled by the combined object is given as:

$d=v\cdot t$

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{d}=\left(1.142\mathrm{m}/\mathrm{s}\right)\left(0.670\mathrm{s}\right) \\ \mathrm{d}=0.946\mathrm{m} \end{gathered}$$

Therefore, the horizontal distance travelled by the combined object is $0.946\mathrm{m}$.