Answer:
[tex]d = \frac{M^2v^2}{2m^2\mu_k g}[/tex]
Explanation:
As we know that there is no external force on skater and the stone so the total momentum of the system will remains constant
so we will have
[tex]Mv + mv_1 = 0[/tex]
here we have
[tex]v_1 = -\frac{M}{m} v[/tex]
so the skater will move back with above speed
now the deceleration of the skater is due to friction given as
[tex]a = \frac{F_f}{m}[/tex]
[tex]a = -\mu_k g[/tex]
so here we have
[tex]v_f^2 - v_i^2 = 2ad[/tex]
[tex]0 - (-\frac{M}{m}v)^2 = 2(-\mu_k g)d[/tex]
[tex]d = \frac{M^2v^2}{2m^2\mu_k g}[/tex]
The distance over which the skater will move in the opposite direction is:
( M² v² ) / ( 2 μk g m² )
[tex]\texttt{ }[/tex]
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
[tex]\large {\boxed {F = ma }[/tex]
F = Force ( Newton )
m = Object's Mass ( kg )
a = Acceleration ( m )
Let us now tackle the problem !
[tex]\texttt{ }[/tex]
Given:
mass of stone = M
speed of stone = v
mass of skater = m
coefficient of kinetic friction = μk
Asked:
distance of skater = d = ?
Solution:
Finally , we will use Conservation of Momentum Law as follows:
[tex]m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2[/tex]
[tex]0 = Mv + mv_2[/tex]
[tex]Mv = -mv_2[/tex]
[tex]\boxed{v_2 = -\frac{M}{m} v}[/tex] → Equation A
[tex]\texttt{ }[/tex]
Next , we will calculate the distance of the skater:
[tex]\Sigma F = ma[/tex]
[tex]-f = ma[/tex]
[tex]-\mu_kN = ma[/tex]
[tex]-\mu_k mg = ma[/tex]
[tex]\boxed{a = -\mu_k g}[/tex] → Equation B
[tex]\texttt{ }[/tex]
[tex]v^2 = (v_2)^2 + 2ad[/tex]
[tex]0^2 = (-\frac{M}{m} v)^2 + 2(-\mu_k g)d[/tex] ← Equation A & Equation B
[tex]d = (\frac{M}{m} v)^2 \div ( 2 \mu_k g)[/tex]
[tex]\boxed{ d = \frac{M^2 v^2}{2 \mu_k g m^2}}[/tex]
[tex]\texttt{ }[/tex]
[tex]\texttt{ }[/tex]
Grade: High School
Subject: Physics
Chapter: Dynamics
