# Problem 8-86 Inelastic collision - Part 3B

A large ball of modeling clay (mass 4.5x102 g) is rolled on a tabletop so that it collides with a stationary small wooden box (mass 7.9x102 g). The collision is completely inelastic, and the ball and box then slide on the table for a distance of 5.1 cm. If the speed of the ball is 2.2 m/s just before the collision, determine: (a) the speed of the ball and box just after the collision (b) the magnitude of the friction force acting on the ball and box.

Accumulated Solution $m_1v_1 + m_2v_2 = (m_1 + m_2)v' \quad \; \text{where}\; v_2 = 0 \\ v{'} = 0.80\; m/s\; \text{ (answer to part (a))}$

$v{'}= \frac{m_1}{m_1 + m_2} v_1 = \frac{4.5 \times 10^2}{12.4 \times 10^2}2.2= 0.798\; m/s = 0.80\; m/s$

Now for part 2 of the problem we simply have an object of mass $m = 1.24\; kg$ with an initial velocity of $v_0 = 0.798\; m/s$ slowing down because of the force of friction. The acceleration of the mass is given by:

(A)   $v = v_0 + at$

(B)   $x = v_{0^t} + (1/2)at^2$

(C)   $v^2 = v_0{^2} + 2ax$