- #1
decentfellow
- 130
- 1
Homework Statement
Board ##A## is placed on board ##B## as shown. Both boards slide, without moving w.r.t each other, along a frictionless horizontal surface at a speed of ##6 \text{m/s}##. Board ##B## hits a resulting board ##C##, "head-on". After the collision, board ##B## and ##C## stick together and the board ##A## slides on top of board ##C## and stops its motion relative to board ##C## in the position shown in the diagram. What is the length (in m) of each board? All three boards have the same mass, size and shape. The coefficient of kinetic friction between boards ##A## & ##B## and b/w ##A## & ##B## is ##0.3##.
Homework Equations
Impulse-momentum theorem:- ##\displaystyle\int{F_{net}dt}=\int{dv}##
##\vec{F}=m\vec{a}##
##f_{max}=\mu N \qquad\qquad\qquad\qquad \text{where, $N$ is the normal reaction acting on the body}##
The Attempt at a Solution
Applying the Impulse momentum theorem on all the blocks we get
$$-\int_{0}^{t}{\mu_km_Agdt}=m_A\int_{6}^{v}{dv}\implies \mu_km_Agt=m_A(6-v)\tag{1}$$
$$-\int_{0}^{t}{(N_3-\mu_km_Ag)dt}=m_A\int_{6}^{v}{dv}\implies (N_3-\mu_km_Ag)t=m_B(6-v)\tag{2}$$
$$\int_{0}^{t}{(N_3+\mu_km_Ag)dt}=m_A\int_{0}^{v}{dv}\implies (N_3+\mu_km_Ag)t=m_Cv\tag{3}$$
From ##(1), (2)## and ##(3)##, we get
$$\int_{0}^{t}{N_3dt}=2\int_{0}^{t}{\mu_km_Agdt}\implies N_3=2\mu_km_Ag$$
$$v=\dfrac{9}{2}\text{m/s}$$
$$t=\dfrac{1}{2}\text{sec}$$
$$a_{A/C}=4\mu_kg$$
$${v_{A/C}}_i=6\text{m/s} \qquad\&\qquad {v_{A/C}}_f=0$$
Now, as per the equation of motion for constant acceleration, we have
$$S_{A/C}=\dfrac{-u^2_{A/C}}{2a_{A/C}}=\dfrac{6^2}{8\mu_kg}=\dfrac{6^2}{4\times 3\times 2}=\dfrac{3}{2}\text{m}$$
But, the answer given in the book is ##10m##, where am I going wrong.
