Answer to Question #142716 in Mechanics | Relativity for Hayri

Question

A 12.0 g of sticky clay is hurled horizontally at a 100-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 the block and the surface is 0.650, what was the speed of the clay immediately before impact?

Answer 1

According to impulse conservation law:

$m_{c}\times V_{c}=m_{b}\times V_{b}$

where on the left sides are mass and initial velocity of clay, and on the right side sum of masses block+clay and their overall velocity after impact.

Also after impact block and clay spent their overall kinetic energy to make work against friction force, so:

$\frac{m_b\times V_{b}^2}{2}=F_{fr}\times l$

where $F_{fr}$ stands for friction force and l is distance.

Friction force finding:

$F_{fr} = k\times m_{b}\times g$

where k - koefficient of friction, g - gravity acceleration.

Calculating:

$F_{fr} = 0.65\times (0.1+0.012)\times 9.8=0.71N$

$V_{b}=\sqrt\frac{2\times F_{fr}\times l}{m_{b}}=\sqrt\frac{2\times 0.71\times 7.5}{0.112}=9.75m/s$

$V_{c}=\frac{m_{b}\times V_{b}}{m_{c}}=\frac{0.112\times 9.75}{0.012}=91m/s$